IB Analysis and Approaches HL · Discrete random variables and the binomial distribution · End of topic

IB exam questions: topic 15

60 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

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Probability distribution of a discrete variable

Exercise 1 BasicIB · P1 TZ2 May 2008

The probability distribution of a discrete random variable $X$ is given by $P (X = x) = c x (5 - x)$ , $x = 1; 2; 3; 4$

a) Find the value of $c$ .

b) Find $E (X)$ .

Show answer

Answer: a) $c = \frac{1}{20}$ ; b) $E (X) = 2.5$

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a) The sum of the probabilities is 1: $c [1 \cdot 4 + 2 \cdot 3 + 3 \cdot 2 + 4 \cdot 1] = c (4 + 6 + 6 + 4) = 20 c = 1 \Rightarrow c = \frac{1}{20} .$

b) The probabilities are $\frac{4}{20}, \frac{6}{20}, \frac{6}{20}, \frac{4}{20}$ for $x = 1, 2, 3, 4$ . $E (X) = \frac{1 \cdot 4 + 2 \cdot 6 + 3 \cdot 6 + 4 \cdot 4}{20} = \frac{50}{20} = 2.5.$

Answer: a) $c = \frac{1}{20}$ ; b) $E (X) = 2.5$

Exercise 2 BasicIB · P1 TZ0 November 2016

The faces of a fair six-sided die are numbered 1, 2, 2, 4, 4, 6. Let $X$ be the discrete random variable that models the score obtained when the die is rolled.

a) Complete the following table for the probability distribution of $X$ .

$x$
$P (X = x)$

b) Find the expected value of $X$ .

Show answer

Answer: a) $P (X = 1) = \frac{1}{6}$ , $P (X = 2) = \frac{2}{6}$ , $P (X = 4) = \frac{2}{6}$ , $P (X = 6) = \frac{1}{6}$ ; b) $E (X) = \frac{19}{6}$

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a) The six faces are 1, 2, 2, 4, 4, 6, all with probability $\frac{1}{6}$ .

$x$	1	2	4	6
$P (X = x)$	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{2}{6}$	$\frac{1}{6}$

b) $E (X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{2}{6} + 4 \cdot \frac{2}{6} + 6 \cdot \frac{1}{6} = \frac{1 + 4 + 8 + 6}{6} = \frac{19}{6} \approx 3.17.$

Answer: a) $P (X = 1) = \frac{1}{6}$ , $P (X = 2) = \frac{2}{6}$ , $P (X = 4) = \frac{2}{6}$ , $P (X = 6) = \frac{1}{6}$ ; b) $E (X) = \frac{19}{6}$

Exercise 3 MediumIB · P1 TZ2 May 2005

The following table shows the probability distribution of a discrete random variable $X$ .

$x$	0	1	2	3
$P (X = x)$	$\frac{1}{5}$	$a$	$b$	$\frac{1}{4}$

a) Given that $E (X) = 1.55$ , find the values of $a$ and $b$ .

b) Calculate $Var (X)$ .

Show answer

Answer: a) $a = 0.3$ ; $b = 0.25$ ; b) $Var (X) = 1.1475 \approx 1.15$

Show working and answer

a) The sum of the probabilities equals 1: $\frac{1}{5} + a + b + \frac{1}{4} = 1 \Rightarrow a + b = \frac{11}{20} = 0.55.$ The expected value is $E (X) = 0 \cdot \frac{1}{5} + 1 \cdot a + 2 \cdot b + 3 \cdot \frac{1}{4} = 1.55$ , that is, $a + 2 b = 0.8$ .

Subtracting the two equations: $b = 0.8 - 0.55 = 0.25$ and so $a = 0.55 - 0.25 = 0.3$ .

b) $E (X^{2}) = 0^{2} \cdot \frac{1}{5} + 1^{2} \cdot 0.3 + 2^{2} \cdot 0.25 + 3^{2} \cdot \frac{1}{4} = 0.3 + 1 + 2.25 = 3.55$ .

$Var (X) = E (X^{2}) - [E (X)]^{2} = 3.55 - 1.5 5^{2} = 1.1475 \approx 1.15.$

Answer: a) $a = 0.3$ ; $b = 0.25$ ; b) $Var (X) = 1.1475 \approx 1.15$

Exercise 4 MediumIB · P1 TZ2 May 2018

The discrete random variable $X$ has the following probability distribution, where $p$ is a constant:

$x$	0	1	2	3	4
$P (X = x)$	$p$	$0.5 - p$	$0.25$	$0.125$	$p^{3}$

a) Find the value of $p$ .

b) i) Find $μ$ , the expected value of $X$ .

(ii) Find $P (X > μ)$ .

Show answer

Answer: a) $p = 0.5$ ; b) i) $μ = 1.375$ ; ii) $P (X > μ) = 0.5$

Show working and answer

a) The sum of the probabilities equals 1: $p + (0.5 - p) + 0.25 + 0.125 + p^{3} = 1 \Rightarrow 0.875 + p^{3} = 1 \Rightarrow p^{3} = 0.125 \Rightarrow p = 0.5.$

The table becomes: $P (X = 0) = 0.5$ , $P (X = 1) = 0$ , $P (X = 2) = 0.25$ , $P (X = 3) = 0.125$ , $P (X = 4) = 0.125$ .

b) (i) $μ = 0 \cdot 0.5 + 1 \cdot 0 + 2 \cdot 0.25 + 3 \cdot 0.125 + 4 \cdot 0.125 = 0.5 + 0.375 + 0.5 = 1.375$ .

(ii) $P (X > 1.375) = P (X = 2) + P (X = 3) + P (X = 4) = 0.25 + 0.125 + 0.125 = 0.5$ .

Answer: a) $p = 0.5$ ; b) i) $μ = 1.375$ ; ii) $P (X > μ) = 0.5$

Exercise 5 MediumIB · P1 TZ1 May 2018

Two fair four-sided (tetrahedral) dice, with faces labelled 1, 2, 3, 4, are rolled and the recorded scores are observed. Let $T$ be the random variable “the maximum of these two scores”. The probability distribution of $T$ is given in the following table.

$t$	1	2	3	4
$P (T = t)$	$\frac{1}{16}$	$a$	$b$	$\frac{7}{16}$

a) Calculate the value of $a$ and the value of $b$ .

b) Calculate the expected value of $T$ .

Show answer

Answer: a) $a = \frac{3}{16}$ , $b = \frac{5}{16}$ ; b) $E (T) = \frac{25}{8}$

Show working and answer

a) There are $4 \cdot 4 = 16$ equally likely outcomes. The maximum is 2 in $(1, 2), (2, 1), (2, 2)$ : $a = \frac{3}{16}$ . Since the probabilities sum to 1: $\frac{1}{16} + \frac{3}{16} + b + \frac{7}{16} = 1 \Rightarrow b = \frac{5}{16} .$ (Check: the maximum is 3 in $(1, 3), (2, 3), (3, 1), (3, 2), (3, 3)$ , that is, 5 cases.)

b) $E (T) = 1 \cdot \frac{1}{16} + 2 \cdot \frac{3}{16} + 3 \cdot \frac{5}{16} + 4 \cdot \frac{7}{16} = \frac{1 + 6 + 15 + 28}{16} = \frac{50}{16} = \frac{25}{8} .$

Answer: a) $a = \frac{3}{16}$ , $b = \frac{5}{16}$ ; b) $E (T) = \frac{25}{8}$

Exercise 6 MediumIB · P1 May 2000

The probability distribution of a discrete random variable $X$ is given by $P (X = x) = k (\frac{2}{3})^{x}, x = 0, 1, 2, \dots$

Calculate the value of $k$ .

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Answer: $k = \frac{1}{3}$

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The sum of all the probabilities must equal 1. It is a geometric series with first term $k$ and common ratio $r = \frac{2}{3}$ (with $∣ r ∣ < 1$ ): $x = 0 \sum \infty k (\frac{2}{3})^{x} = \frac{k}{1 - \frac{2}{3}} = 3 k = 1 \Rightarrow k = \frac{1}{3} .$

Answer: $k = \frac{1}{3}$

Exercise 7 MediumIB · P1 May 1998

The discrete random variable $X$ has the following probability distribution.

$x$	0	1	2	3	4
$P (X = x)$	$\frac{1}{8}$	$3 k$	$\frac{1}{6} k$	$\frac{1}{4}$	$\frac{1}{6} k$

a) Calculate the exact value of $k$ .

b) Calculate $P (0 < X < 4)$ .

Show answer

Answer: a) $k = \frac{3}{16}$ ; b) $P (0 < X < 4) = \frac{27}{32}$

Show working and answer

a) The sum of the probabilities is 1: $\frac{1}{8} + 3 k + \frac{k}{6} + \frac{1}{4} + \frac{k}{6} = 1 \Rightarrow \frac{10 k}{3} = \frac{5}{8} \Rightarrow k = \frac{3}{16} .$ (We use $3 k + \frac{k}{6} + \frac{k}{6} = \frac{10 k}{3}$ and $1 - \frac{1}{8} - \frac{1}{4} = \frac{5}{8}$ .)

b) $P (0 < X < 4) = P (X = 1) + P (X = 2) + P (X = 3) = 3 k + \frac{k}{6} + \frac{1}{4} = \frac{9}{16} + \frac{1}{32} + \frac{1}{4} = \frac{18 + 1 + 8}{32} = \frac{27}{32} .$

Answer: a) $k = \frac{3}{16}$ ; b) $P (0 < X < 4) = \frac{27}{32}$

Exercise 8 MediumIB · P1 November 1998

The discrete random variable $X$ has the probability distribution shown in the table below.

$x$	0	1	2	3
$P (X = x)$	$k$	$\frac{1}{4}$	$\frac{1}{4}$	$3 k$

a) Calculate the value of $k$ .

b) Calculate $E (X)$ .

Show answer

Answer: a) $k = \frac{1}{8}$ ; b) $E (X) = \frac{15}{8}$

Show working and answer

a) The sum of the probabilities is 1: $k + \frac{1}{4} + \frac{1}{4} + 3 k = 1 \Rightarrow 4 k = \frac{1}{2} \Rightarrow k = \frac{1}{8} .$

b) $E (X) = \sum x P (X = x)$ : $E (X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{3}{8} = \frac{1}{4} + \frac{1}{2} + \frac{9}{8} = \frac{15}{8} .$

Answer: a) $k = \frac{1}{8}$ ; b) $E (X) = \frac{15}{8}$

Exercise 9 MediumIB · P1 May 2000

In a game a player rolls a biased tetrahedral (four-sided) die. The probability of each possible score is shown below.

Score	1	2	3	4
Probability	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{1}{10}$	$x$

Calculate the probability of a total score of six after two rounds.

Show answer

Answer: $\frac{1}{4}$ (with $x = \frac{3}{10}$ )

Show working and answer

The probabilities sum to 1: $x = 1 - (\frac{1}{5} + \frac{2}{5} + \frac{1}{10}) = 1 - \frac{7}{10} = \frac{3}{10} .$
To obtain a total of 6 in two rounds, the possible pairs are $(2, 4)$ , $(4, 2)$ and $(3, 3)$ (with $(1, 5)$ and $(5, 1)$ impossible, since there are only scores from 1 to 4): $P = 2 \cdot \frac{2}{5} \cdot \frac{3}{10} + (\frac{1}{10})^{2} = \frac{12}{50} + \frac{1}{100} = \frac{25}{100} = \frac{1}{4} .$

Answer: $\frac{1}{4}$ (with $x = \frac{3}{10}$ )

Exercise 10 MediumIB · P1 TZ2 May 2004

The discrete random variable $X$ has the following probability distribution: $P (X = x) = ⎩ ⎨ ⎧ \frac{k}{x}, 0, x = 1, 2, 3, 4 otherwise$

Calculate

a) the value of the constant $k$ .

b) $E (X)$ .

Show answer

Answer: a) $k = \frac{12}{25}$ ; b) $E (X) = \frac{48}{25}$

Show working and answer

a) The probabilities sum to 1: $k (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = k \cdot \frac{25}{12} = 1 \Rightarrow k = \frac{12}{25} .$

b) $E (X) = \sum x \cdot \frac{k}{x} = k + k + k + k = 4 k = \frac{48}{25}$ .

Answer: a) $k = \frac{12}{25}$ ; b) $E (X) = \frac{48}{25}$

Exercise 11 MediumIB · P1 TZ2 May 2014

A pack of seven transistors contains three that are defective. Three transistors are chosen at random from the pack, without replacement. The discrete random variable $X$ represents the number of defective transistors that have been chosen.

(i) Find $P (X = 2)$ .

(ii) Copy and complete the following table.

$x$	0	1	2	3
$P (X = x)$

(iii) Determine $E (X)$ .

Show answer

Answer: i) $\frac{12}{35}$ ; ii) $\frac{4}{35}, \frac{18}{35}, \frac{12}{35}, \frac{1}{35}$ ; iii) $E (X) = \frac{9}{7}$

Show working and answer

There are 7 transistors: 3 defective and 4 non-defective. Three are chosen without replacement, giving $(3 7) = 35$ possible selections.

(i) Two defective and one non-defective: $P (X = 2) = \frac{( 2 3 ) ( 1 4 )}{35} = \frac{12}{35} .$

(ii) $P (X = 0) = \frac{( 3 4 )}{35} = \frac{4}{35}, P (X = 1) = \frac{( 1 3 ) ( 2 4 )}{35} = \frac{18}{35}, P (X = 3) = \frac{( 3 3 )}{35} = \frac{1}{35} .$

$x$	0	1	2	3
$P (X = x)$	$\frac{4}{35}$	$\frac{18}{35}$	$\frac{12}{35}$	$\frac{1}{35}$

(iii) $E (X) = \frac{0 \cdot 4 + 1 \cdot 18 + 2 \cdot 12 + 3 \cdot 1}{35} = \frac{45}{35} = \frac{9}{7} .$

Answer: i) $\frac{12}{35}$ ; ii) $\frac{4}{35}, \frac{18}{35}, \frac{12}{35}, \frac{1}{35}$ ; iii) $E (X) = \frac{9}{7}$

Exercise 12 MediumIB · P2 TZ1 May 2022

A discrete random variable, $X$ , has the following probability distribution:

$x$	0	1	2	3
$P (X = x)$	$0.41$	$k - 0.28$	$0.46$	$0.29 - 2 k^{2}$

a) Show that $2 k^{2} - k + 0.12 = 0$ .

b) Calculate the value of $k$ , giving a reason for your answer.

c) Hence, calculate $E (X)$ .

Show answer

Answer: a) shown; b) $k = 0.3$ ; c) $E (X) = 1.27$

Show working and answer

a) The sum of the probabilities is 1: $0.41 + (k - 0.28) + 0.46 + (0.29 - 2 k^{2}) = 1 \Rightarrow 0.88 + k - 2 k^{2} = 1 \Rightarrow 2 k^{2} - k + 0.12 = 0.$

b) Using the GDC or the formula: $k = \frac{1 \pm 1 - 0.96}{4} = \frac{1 \pm 0.2}{4} \Rightarrow k = 0.3 or k = 0.2.$ If $k = 0.2$ , then $P (X = 1) = k - 0.28 = - 0.08 < 0$ , which is impossible for a probability. Therefore $k = 0.3$ .

c) With $k = 0.3$ : $P (X = 1) = 0.02$ and $P (X = 3) = 0.29 - 2 (0.09) = 0.11$ . $E (X) = 0 \cdot 0.41 + 1 \cdot 0.02 + 2 \cdot 0.46 + 3 \cdot 0.11 = 0.02 + 0.92 + 0.33 = 1.27.$

Answer: a) shown; b) $k = 0.3$ ; c) $E (X) = 1.27$

Exercise 13 HardIB · P2 TZ0 November 2010

Tim rolls two fair, identical dice simultaneously. Each die has six faces. Two faces show the number 1, two faces show the number 2 and two faces show the number 3. Tim's score is the sum of the two numbers obtained with the dice.

a) (i) Calculate the probability that Tim's score is equal to 6.

(ii) Calculate the probability that Tim's score is at least 3.

Tim plays with his friend Hill, who also has the two dice, numbered in the same way. Hill's score is the sum of the two numbers obtained with his dice.

b) (i) Calculate the probability that Tim and Hill both obtain a score equal to 6.

(ii) Calculate the probability that Tim and Hill obtain the same score.

c) Let $X$ be the highest of the numbers obtained with the four dice.

(i) Verify that $P (X \leq 2) = \frac{16}{81}$ .

(ii) Complete the following probability distribution table.

$X$	1	2	3
$P (X = x)$	$\frac{1}{81}$

(iii) Calculate $E (X)$ and $E (X^{2})$ and, hence, find $Var (X)$ .

d) Given that $X = 3$ , find the probability that the sum of the numbers on the four dice is equal to 8.

Show answer

Answer: a) (i) $\frac{1}{9}$ ; (ii) $\frac{8}{9}$ ; b) (i) $\frac{1}{81}$ ; (ii) $\frac{19}{81}$ ; c) $P (X = 2) = \frac{15}{81}$ , $P (X = 3) = \frac{65}{81}$ , $E (X) = \frac{226}{81} \approx 2.79$ , $E (X^{2}) = \frac{646}{81} \approx 7.98$ , $Var (X) = \frac{1250}{6561} \approx 0.191$ ; d) $\frac{18}{65}$

Show working and answer

Each die gives 1, 2 or 3, each with probability $\frac{1}{3}$ (two faces out of six).

a) Tim's score is the sum of two dice. The probabilities of each sum are:

$t$	2	3	4	5	6
$P (T = t)$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{3}{9}$	$\frac{2}{9}$	$\frac{1}{9}$

(i) $P (T = 6) = P (3, 3) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$ .

(ii) $P (T \geq 3) = 1 - P (T = 2) = 1 - \frac{1}{9} = \frac{8}{9}$ .

b) (i) Hill's dice are independent of Tim's: $P (T = 6, H = 6) = \frac{1}{9} \cdot \frac{1}{9} = \frac{1}{81}$ .

(ii) Same score: $\sum_{t} P (T = t)^{2} = \frac{1 ^{2} + 2 ^{2} + 3 ^{2} + 2 ^{2} + 1 ^{2}}{81} = \frac{19}{81}$ .

c) (i) $X \leq 2$ means that the four dice show 1 or 2, each with probability $\frac{2}{3}$ : $P (X \leq 2) = (\frac{2}{3})^{4} = \frac{16}{81} .$

(ii) $P (X = 1) = (\frac{1}{3})^{4} = \frac{1}{81}$ . Then $P (X = 2) = P (X \leq 2) - P (X = 1) = \frac{16}{81} - \frac{1}{81} = \frac{15}{81}$ and $P (X = 3) = 1 - \frac{16}{81} = \frac{65}{81}$ .

$X$	1	2	3
$P (X = x)$	$\frac{1}{81}$	$\frac{15}{81}$	$\frac{65}{81}$

(iii) $E (X) = \frac{1 + 2 \cdot 15 + 3 \cdot 65}{81} = \frac{226}{81} \approx 2.79$ and $E (X^{2}) = \frac{1 + 4 \cdot 15 + 9 \cdot 65}{81} = \frac{646}{81} \approx 7.98$ .

$Var (X) = \frac{646}{81} - (\frac{226}{81})^{2} = \frac{52326 - 51076}{6561} = \frac{1250}{6561} \approx 0.191.$

d) The four values (1, 2 or 3) are equally likely on each die, so there are $3^{4} = 81$ equally likely combinations. The cases with $X = 3$ number $81 - 16 = 65$ . Among them, those whose sum is 8 and that contain at least one 3 are the arrangements of ${3, 3, 1, 1}$ (6) and of ${3, 2, 2, 1}$ (12), that is, 18 cases. (The combination ${2, 2, 2, 2}$ also sums to 8, but it has no 3.) $P (sum = 8 ∣ X = 3) = \frac{18/81}{65/81} = \frac{18}{65} .$

Exercise 14 HardIB · P1 TZ2 May 2012

Consider the following functions: $f (x) = \frac{2 x ^{2} + 3}{75}$ , $x \geq 0$ \qquad $g (x) = \frac{∣3 x - 4∣}{10}$ , $x \in R$ .

a) State the range of $f$ and the range of $g$ .

b) Find an expression for the composite function $(f \circ g) (x)$ , in the form $\frac{a x ^{2} + b x + c}{3750}$ , where $a, b, c \in Z$ .

c) i) Find an expression for the inverse function $f^{- 1} (x)$ .

(ii) State the domain and range of $f^{- 1}$ .

Now the domains of $f$ and $g$ are restricted to ${0, 1, 2, 3, 4}$ .

d) Considering the values of $f$ and $g$ on this domain, determine which of the two functions $f$ and $g$ could be used to find a probability distribution for a discrete random variable $X$ . Explain clearly the reasons for your answer.

e) Using this probability distribution, calculate the mean of $X$ .

Show answer

Answer: a) $R_{f} = [\frac{1}{25}, + \infty [$ , $R_{g} = [0, + \infty [$ ; b) $(f \circ g) (x) = \frac{9 x ^{2} - 24 x + 166}{3750}$ ; c) i) $f^{- 1} (x) = \frac{75 x - 3}{2}$ , ii) $D_{f^{- 1}} = [\frac{1}{25}, + \infty [$ , $R_{f^{- 1}} = [0, + \infty [$ ; d) $f$ (the sum is 1; that of $g$ is 2); e) $E (X) = \frac{46}{15} \approx 3.07$

Show working and answer

a) For $x \geq 0$ : $x^{2} \geq 0$ , so $f (x) \geq \frac{3}{75} = \frac{1}{25}$ and $f$ increases without bound: $R_{f} = [\frac{1}{25}, + \infty [$ . The absolute value is always $\geq 0$ and $3 x - 4$ takes any real value: $R_{g} = [0, + \infty [$ .

b) $(f \circ g) (x) = f (\frac{∣3 x - 4∣}{10}) = \frac{2 \cdot \frac{( 3 x - 4 ) ^{2}}{100} + 3}{75} = \frac{2 ( 9 x ^{2} - 24 x + 16 ) + 300}{7500} = \frac{9 x ^{2} - 24 x + 166}{3750} .$

c) i) Let $y = \frac{2 x ^{2} + 3}{75}$ . Solving for $x$ , $x^{2} = \frac{75 y - 3}{2}$ and, since $x \geq 0$ , $x = \frac{75 y - 3}{2}$ . Hence $f^{- 1} (x) = \frac{75 x - 3}{2} .$

(ii) The domain of $f^{- 1}$ is the range of $f$ and its range is the domain of $f$ : $D_{f^{- 1}} = [\frac{1}{25}, + \infty [$ and $R_{f^{- 1}} = [0, + \infty [$ .

d) With domain ${0, 1, 2, 3, 4}$ :

$x$	0	1	2	3	4
$f (x)$	$\frac{1}{25}$	$\frac{1}{15}$	$\frac{11}{75}$	$\frac{7}{25}$	$\frac{7}{15}$
$g (x)$	$\frac{4}{10}$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{5}{10}$	$\frac{8}{10}$

A probability distribution requires values between 0 and 1 whose sum is 1. $\sum f (x) = \frac{3 + 5 + 11 + 21 + 35}{75} = 1$ , whereas $\sum g (x) = 2 \neq = 1$ . Hence only $f$ can be used.

e) $E (X) = \frac{0 \cdot 3 + 1 \cdot 5 + 2 \cdot 11 + 3 \cdot 21 + 4 \cdot 35}{75} = \frac{230}{75} = \frac{46}{15} \approx 3.07.$

Exercise 15 HardIB · P2 TZ0 November 2016

A chocolate shop advertises free gifts for customers who collect three coupons. Coupons are placed at random in 10% of the chocolate bars sold in the shop. Kati buys some of these chocolate bars and opens them one by one to see whether they contain a coupon. Let $P (X = n)$ be the probability that Kati obtains her third coupon in the $n$ th chocolate bar she opens. (Assume that the probability that a given chocolate bar contains a coupon remains 10% throughout the question.)

a) Show that $P (X = 3) = 0.001$ and $P (X = 4) = 0.0027$ .

It is known that $P (X = n) = \frac{n ^{2} + an + b}{2000} \times 0. 9^{n - 3} for n \geq 3, n \in N .$

b) Find the values of the constants $a$ and $b$ .

c) Deduce that $\frac{P ( X = n )}{P ( X = n - 1 )} = \frac{0.9 ( n - 1 )}{n - 3} for n > 3.$

d) (i) Hence show that $X$ has two modes, $m_{1}$ and $m_{2}$ .

(ii) State the values of $m_{1}$ and $m_{2}$ .

Kati's mother goes to the shop and buys $x$ chocolate bars. She takes them home for Kati to open.

e) Determine the minimum value of $x$ such that the probability that Kati receives at least one free gift is greater than $0.5$ .

Show answer

Answer: a) $P (X = 3) = 0.001$ , $P (X = 4) = 0.0027$ ; b) $a = - 3$ , $b = 2$ ; d) ii) $m_{1} = 20$ , $m_{2} = 21$ ; e) $x = 27$

Show working and answer

Kati opens chocolate bars; each one contains a coupon with probability $0.1$ . $X = n$ means that the third coupon appears exactly in the $n$ th bar.

a) For $n = 3$ : all three contain a coupon: $0. 1^{3} = 0.001$ . For $n = 4$ : the fourth contains a coupon and exactly two of the first three do: $P (X = 4) = (2 3) 0. 1^{2} \cdot 0.9 \cdot 0.1 = 3 \cdot 0. 1^{3} \cdot 0.9 = 0.0027.$

b) We use the given formula.

$n = 3$ : $\frac{9 + 3 a + b}{2000} \cdot 0. 9^{0} = 0.001 \Rightarrow 3 a + b = - 7$ .
$n = 4$ : $\frac{16 + 4 a + b}{2000} \cdot 0.9 = 0.0027 \Rightarrow 16 + 4 a + b = 6 \Rightarrow 4 a + b = - 10$ .

Subtracting: $a = - 3$ and $b = 2$ . (Thus $n^{2} - 3 n + 2 = (n - 1) (n - 2)$ .)

c) With $a = - 3$ , $b = 2$ : $P (X = n) = \frac{( n - 1 ) ( n - 2 )}{2000} 0. 9^{n - 3}$ . For $n > 3$ : $\frac{P ( X = n )}{P ( X = n - 1 )} = \frac{( n - 1 ) ( n - 2 ) 0. 9 ^{n - 3}}{( n - 2 ) ( n - 3 ) 0. 9 ^{n - 4}} = \frac{0.9 ( n - 1 )}{n - 3} .$

d) (i) The ratio equals 1 when $0.9 (n - 1) = n - 3$ , that is, $0.1 n = 2.1$ , so $n = 21$ . Moreover, the ratio is greater than 1 if $n < 21$ and less than 1 if $n > 21$ . Hence $P (X = n)$ increases up to $n = 20$ , $P (X = 21) = P (X = 20)$ , and then decreases: there are two modes.

(ii) $m_{1} = 20$ and $m_{2} = 21$ .

e) Kati's mother buys $x$ bars. There is a free gift if at least 3 coupons appear. Let $C \sim B (x; 0.1)$ be the number of coupons: we need $P (C \geq 3) > 0.5$ . Using the GDC (binomial distribution, cumulative probability):

$x$	26	27
$P (C \geq 3)$	$0.4895$	$0.5154$

The minimum value is $x = 27$ .

Answer: a) $P (X = 3) = 0.001$ , $P (X = 4) = 0.0027$ ; b) $a = - 3$ , $b = 2$ ; d) ii) $m_{1} = 20$ , $m_{2} = 21$ ; e) $x = 27$

Exercise 16 HardIB · P2 May 2001

In a game, the probability that a player scores with a shot is $\frac{1}{4}$ . Let $X$ be the number of shots the player needs to score, including shots from any previous scoring attempts. (You may assume that each shot is independent of the previous ones.)

a) Calculate $P (X = 3)$ .

b) Calculate the probability that the player has at least three misses before scoring twice.

c) Show that the expected value of $X$ is $4$ .

(You may use the result $(1 - x)^{- 2} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots$ )

Show answer

Answer: a) $\frac{9}{64}$ ; b) $\frac{189}{256}$ ; c) $E (X) = 4$

Show working and answer

$X$ is the number of shots up to and including the first successful shot: $P (X = x) = (\frac{3}{4})^{x - 1} \frac{1}{4}$ .

a) Two misses and then a score: $P (X = 3) = (\frac{3}{4})^{2} \cdot \frac{1}{4} = \frac{9}{64} .$

b) "At least three misses before scoring twice" is the complement of "scoring twice in 4 shots or fewer", that is, of having at least 2 scores in 4 shots. Let $Y \sim B (4; \frac{1}{4})$ : $P (Y \geq 2) = 1 - (\frac{3}{4})^{4} - 4 \cdot \frac{1}{4} (\frac{3}{4})^{3} = 1 - \frac{81}{256} - \frac{108}{256} = \frac{67}{256} .$ Hence the required probability is $1 - \frac{67}{256} = \frac{189}{256} .$

c) With $q = \frac{3}{4}$ : $E (X) = x = 1 \sum \infty x q^{x - 1} \cdot \frac{1}{4} = \frac{1}{4} (1 + 2 q + 3 q^{2} + 4 q^{3} + \dots) = \frac{1}{4} (1 - q)^{- 2} .$ Since $1 - q = \frac{1}{4}$ , we get $E (X) = \frac{1}{4} \cdot 16 = 4$ .

Answer: a) $\frac{9}{64}$ ; b) $\frac{189}{256}$ ; c) $E (X) = 4$

Exercise 17 HardIB · P1 TZ1 May 2021

A four-sided biased die A is rolled. Let $X$ be the score obtained when die A is rolled. The probability distribution for $X$ is shown in the following table.

$x$	1	2	3	4
$P (X = x)$	$p$	$p$	$p$	$\frac{1}{2} p$

a) Find the value of $p$ .

b) Using the above, find the value of $E (X)$ .

A second biased four-sided die, $B$ , is rolled. Let $Y$ be the score obtained when die $B$ is rolled. The probability distribution for $Y$ is shown in the following table.

$y$	1	2	3	4
$P (Y = y)$	$q$	$q$	$q$	$r$

c) (i) Write down the range of possible values of $r$ .

(ii) Using the above, find the range of possible values of $q$ .

d) Using the above, find the range of possible values of $E (Y)$ .

Agnes and Bárbara play a game using these dice. Agnes rolls die A once and Bárbara rolls die B once. The probability that Agnes's score is less than Bárbara's is $\frac{1}{2}$ .

e) Calculate the value of $E (Y)$ .

Show answer

Answer: a) $p = \frac{2}{7}$ ; b) $E (X) = \frac{16}{7}$ ; c) i) $0 \leq r \leq 1$ ; ii) $0 \leq q \leq \frac{1}{3}$ ; d) $2 \leq E (Y) \leq 4$ ; e) $E (Y) = \frac{11}{4}$

Show working and answer

a) $p + p + p + \frac{1}{2} p = 1 \Rightarrow \frac{7}{2} p = 1 \Rightarrow p = \frac{2}{7}$ .

b) $E (X) = (1 + 2 + 3) p + 4 \cdot \frac{1}{2} p = 6 p + 2 p = 8 p = \frac{16}{7} .$

c) (i) $r$ is a probability: $0 \leq r \leq 1$ .

(ii) $3 q + r = 1$ , so $q = \frac{1 - r}{3}$ . With $0 \leq r \leq 1$ : $0 \leq q \leq \frac{1}{3}$ .

d) $E (Y) = (1 + 2 + 3) q + 4 r = 6 q + 4 r = 2 (1 - r) + 4 r = 2 + 2 r .$ With $0 \leq r \leq 1$ : $2 \leq E (Y) \leq 4$ .

e) Agnes scores less than Barbara: $P (X < Y) = P (Y = 2) P (X = 1) + P (Y = 3) P (X \leq 2) + P (Y = 4) P (X \leq 3) = q \cdot \frac{2}{7} + q \cdot \frac{4}{7} + r \cdot \frac{6}{7} = \frac{6}{7} (q + r) .$ Setting this equal to $\frac{1}{2}$ : $q + r = \frac{7}{12}$ . Together with $3 q + r = 1$ , subtracting gives $2 q = \frac{5}{12}$ , so $q = \frac{5}{24}$ and $r = \frac{7}{12} - \frac{5}{24} = \frac{3}{8}$ . Then $E (Y) = 2 + 2 r = 2 + \frac{3}{4} = \frac{11}{4} .$

Answer: a) $p = \frac{2}{7}$ ; b) $E (X) = \frac{16}{7}$ ; c) i) $0 \leq r \leq 1$ ; ii) $0 \leq q \leq \frac{1}{3}$ ; d) $2 \leq E (Y) \leq 4$ ; e) $E (Y) = \frac{11}{4}$

Exercise 18 HardIB · P1 TZ2 May 2022

A four-sided die, which is not fair, with faces labelled 1, 2, 3 and 4, is rolled and the result is recorded. Let $X$ be the result obtained when the die is rolled. The following table shows the probability distribution of $X$ , where $p$ and $q$ are constants.

$x$	1	2	3	4
$P (X = x)$	$p$	$0.3$	$q$	$0.1$

For this probability distribution, it is known that $E (X) = 2$ .

a) Show that $p = 0.4$ and $q = 0.2$ .

b) Find $P (X > 2)$ .

Nicky decides to play a game with this four-sided die. In this game, the die may be rolled at most five times. The score is calculated by adding the results of all the rolls. Nicky wins the game if the score is at least ten.

After rolling the die three times, Nicky's score is equal to four.

c) Assuming that the rolls of the die are independent of each other, find the probability that Nicky wins the game.

David has two pairs of fair four-sided dice: a yellow pair and a red pair. The two yellow dice have their faces labelled 1, 2, 3 and 4. Let $S$ be the sum obtained when the two yellow dice are rolled. The probability distribution of $S$ is shown below.

$s$	2	3	4	5	6	7	8
$P (S = s)$	$\frac{1}{16}$	$\frac{2}{16}$	$\frac{3}{16}$	$\frac{4}{16}$	$\frac{3}{16}$	$\frac{2}{16}$	$\frac{1}{16}$

The first red die has its faces labelled 1, 2, 2 and 3. The second red die has its faces labelled 1, $a$ , $a$ and $b$ , where $a < b$ and $a, b \in Z^{+}$ . The probability distribution of the sum obtained when the red pair is rolled is equal to the probability distribution of the sum obtained when the yellow pair is rolled.

d) Determine the value of $b$ .

e) Find the value of $a$ , giving evidence to support your answer.

Show answer

Answer: a) $p = \frac{2}{5}$ , $q = \frac{1}{5}$ ; b) $P (X > 2) = \frac{3}{10}$ ; c) $\frac{3}{20}$ ; d) $b = 5$ ; e) $a = 3$

Show working and answer

a) The sum of the probabilities is 1: $p + 0.3 + q + 0.1 = 1 \Rightarrow p + q = 0.6$ . The mean is 2: $p + 2 (0.3) + 3 q + 4 (0.1) = 2 \Rightarrow p + 3 q = 1$ . Subtracting: $2 q = 0.4$ , so $q = 0.2$ and $p = 0.4$ .

b) $P (X > 2) = P (X = 3) + P (X = 4) = 0.2 + 0.1 = 0.3$ .

c) With 3 rolls made, the score is 4 and 2 rolls remain, which must add up to at least 6. The possible pairs are $(2, 4)$ , $(4, 2)$ , $(3, 3)$ , $(3, 4)$ , $(4, 3)$ and $(4, 4)$ : $P = 2 (0.3) (0.1) + (0.2)^{2} + 2 (0.2) (0.1) + (0.1)^{2} = 0.06 + 0.04 + 0.04 + 0.01 = 0.15.$

d) The maximum sum of the yellow pair is 8, with probability $\frac{1}{16}$ . For the red pair the maximum sum is $3 + b$ , so $3 + b = 8$ and $b = 5$ .

e) The red dice are ${1, 2, 2, 3}$ and ${1, a, a, 5}$ with $1 < a < 5$ or $a = 1$ .

$P (S = 2) = \frac{1}{16}$ : the sum 2 only comes from $1 + 1$ , with probability $\frac{1}{4} \cdot P (getting 1 on the 2nd die) = \frac{1}{16}$ . So the second die has only one 1, and therefore $a \neq = 1$ .
$P (S = 3) = \frac{2}{16}$ : $2 + 1$ already appears with probability $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} = \frac{2}{16}$ . If $a = 2$ , then $1 + 2$ would also appear and the probability would be greater. So $a \neq = 2$ .
$P (S = 4) = \frac{3}{16}$ : $3 + 1$ contributes $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$ . The remaining $\frac{2}{16}$ can only come from $1 + a$ with $a = 3$ : $\frac{1}{4} \cdot \frac{2}{4} = \frac{2}{16}$ . (With $a = 4$ the total does not match.)

Check with $a = 3$ , $b = 5$ : the sums $2, 3, 4, 5, 6, 7, 8$ have probabilities $\frac{1}{16}, \frac{2}{16}, \frac{3}{16}, \frac{4}{16}, \frac{3}{16}, \frac{2}{16}, \frac{1}{16}$ , the same as those of the yellow pair. Therefore $a = 3$ .

Answer: a) $p = \frac{2}{5}$ , $q = \frac{1}{5}$ ; b) $P (X > 2) = \frac{3}{10}$ ; c) $\frac{3}{20}$ ; d) $b = 5$ ; e) $a = 3$

Expected value, variance and games

Exercise 19 BasicIB · P1 TZ0 November 2004

A perfectly balanced die has six faces showing the numbers 1, 1, 2, 3, 4, 5. Find the mean and variance of the score.

Show answer

Answer: mean $= \frac{8}{3}$ ; variance $= \frac{20}{9}$

Show working and answer

Values $1, 1, 2, 3, 4, 5$ , each face with probability $\frac{1}{6}$ (the 1 appears twice).

Mean: $E (X) = \frac{1 + 1 + 2 + 3 + 4 + 5}{6} = \frac{16}{6} = \frac{8}{3} .$

Variance: $Var (X) = E (X^{2}) - [E (X)]^{2}$ with $E (X^{2}) = \frac{1 + 1 + 4 + 9 + 16 + 25}{6} = \frac{56}{6} = \frac{28}{3},$ $Var (X) = \frac{28}{3} - \frac{64}{9} = \frac{84 - 64}{9} = \frac{20}{9} .$

Answer: mean $= \frac{8}{3}$ ; variance $= \frac{20}{9}$

Exercise 20 MediumIB · P2 TZ2 May 2010

A discrete random variable $X$ has the probability distribution given in the following table.

$X$	0.5	1.5	2.5	3.5	4.5	5.5
$P (X = x)$	0.15	0.21	$p$	$q$	0.13	0.07

a) If $E (X) = 2.61$ , determine the values of $p$ and $q$ .

b) Calculate $Var (X)$ , correct to three significant figures.

Show answer

Answer: a) $p = 0.29$ ; $q = 0.15$ ; b) $Var (X) \approx 2.10$

Show working and answer

a) The sum of the probabilities equals 1: $0.15 + 0.21 + p + q + 0.13 + 0.07 = 1$ , so $p + q = 0.44$ .

The expected value is $E (X) = \sum x P (X = x)$ : $0.5 \cdot 0.15 + 1.5 \cdot 0.21 + 2.5 p + 3.5 q + 4.5 \cdot 0.13 + 5.5 \cdot 0.07 = 2.61.$ The known terms sum to $0.075 + 0.315 + 0.585 + 0.385 = 1.36$ , so $2.5 p + 3.5 q = 1.25$ .

Substituting $q = 0.44 - p$ : $2.5 p + 1.54 - 3.5 p = 1.25$ , hence $p = 0.29$ and $q = 0.15$ .

b) $E (X^{2}) = 0.25 \cdot 0.15 + 2.25 \cdot 0.21 + 6.25 \cdot 0.29 + 12.25 \cdot 0.15 + 20.25 \cdot 0.13 + 30.25 \cdot 0.07 = 8.91$ .

$Var (X) = 8.91 - 2.6 1^{2} = 2.0979 \approx 2.10.$

Answer: a) $p = 0.29$ ; $q = 0.15$ ; b) $Var (X) \approx 2.10$

Exercise 21 MediumIB · P1 TZ0 November 2008

John removes the labels from three tins of tomato soup and two tins of chicken soup to take part in a prize draw, and keeps the tins. At that moment, he realises that the tins are identical, so it is not possible to tell the tomato soup tins from the chicken soup tins. A few weeks later, he decides to have chicken soup for lunch. He opens the tins at random until he opens a tin of chicken. Let $Y$ be the number of tins he opens. Find:

a) the possible values of $Y$ .

b) the probability of each of these values of $Y$ .

c) the expected value of $Y$ .

Show answer

Answer: a) $y = 1, 2, 3, 4$ ; b) $P (Y = 1) = \frac{2}{5}$ , $P (Y = 2) = \frac{3}{10}$ , $P (Y = 3) = \frac{1}{5}$ , $P (Y = 4) = \frac{1}{10}$ ; c) $E (Y) = 2$

Show working and answer

There are 5 tins: 3 of tomato (T) and 2 of chicken (P). They are opened at random, without replacement, until one of chicken comes up.

a) At the earliest, the first tin is chicken ( $Y = 1$ ). At the latest, the 3 tomato tins come out before the first chicken one ( $Y = 4$ ). Therefore $Y = 1, 2, 3, 4$ .

b) We multiply the probabilities along each sequence:

$P (Y = 1) = \frac{2}{5}$ .
$P (Y = 2) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$ .
$P (Y = 3) = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} = \frac{1}{5}$ .
$P (Y = 4) = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} \cdot 1 = \frac{1}{10}$ .

The sum is $\frac{4 + 3 + 2 + 1}{10} = 1$ .

c) $E (Y) = 1 \cdot \frac{4}{10} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{2}{10} + 4 \cdot \frac{1}{10} = \frac{4 + 6 + 6 + 4}{10} = 2.$

Answer: a) $y = 1, 2, 3, 4$ ; b) $P (Y = 1) = \frac{2}{5}$ , $P (Y = 2) = \frac{3}{10}$ , $P (Y = 3) = \frac{1}{5}$ , $P (Y = 4) = \frac{1}{10}$ ; c) $E (Y) = 2$

Exercise 22 MediumIB · P2 TZ0 November 2016

A random variable $X$ has the probability distribution shown in the following table.

$x$	0.5	1.5	2.5	3.5	4.5	5.5
$P (X = x)$	0.12	0.18	0.20	0.28	0.14	0.08

a) Determine the value of $E (X^{2})$ .

b) Find the value of $Var (X)$ .

Show answer

Answer: a) $E (X^{2}) = 10.37$ ; b) $Var (X) \approx 2.08$

Show working and answer

a) $E (X^{2}) = \sum x^{2} P (X = x)$ : $0.25 \cdot 0.12 + 2.25 \cdot 0.18 + 6.25 \cdot 0.20 + 12.25 \cdot 0.28 + 20.25 \cdot 0.14 + 30.25 \cdot 0.08$ $= 0.03 + 0.405 + 1.25 + 3.43 + 2.835 + 2.42 = 10.37.$

b) First, $E (X) = 0.06 + 0.27 + 0.5 + 0.98 + 0.63 + 0.44 = 2.88$ . Then $Var (X) = E (X^{2}) - [E (X)]^{2} = 10.37 - 2.8 8^{2} = 10.37 - 8.2944 = 2.0756 \approx 2.08.$

Answer: a) $E (X^{2}) = 10.37$ ; b) $Var (X) \approx 2.08$

Exercise 23 MediumIB · P1 May 1998

When a biased die is rolled, the numbers 1 to 6 appear according to the following probability distribution.

Number on the upper face of the die	1	2	3	4	5	6
Probability	$\frac{2}{9}$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{9}$

If $X$ is the random variable that takes the values 1 to 6 according to the table above, find the exact values of $E (X)$ and $Var (X)$ .

Show answer

Answer: $E (X) = \frac{10}{3}$ ; $Var (X) = \frac{26}{9}$

Show working and answer

$E (X) = \frac{2 \cdot 1 + 1 \cdot 2 + 2 \cdot 3 + 1 \cdot 4 + 2 \cdot 5 + 1 \cdot 6}{9} = \frac{2 + 2 + 6 + 4 + 10 + 6}{9} = \frac{30}{9} = \frac{10}{3} .$

$E (X^{2}) = \frac{2 \cdot 1 + 1 \cdot 4 + 2 \cdot 9 + 1 \cdot 16 + 2 \cdot 25 + 1 \cdot 36}{9} = \frac{2 + 4 + 18 + 16 + 50 + 36}{9} = \frac{126}{9} = 14.$

$Var (X) = E (X^{2}) - [E (X)]^{2} = 14 - \frac{100}{9} = \frac{26}{9} .$

Answer: $E (X) = \frac{10}{3}$ ; $Var (X) = \frac{26}{9}$

Exercise 24 MediumIB · P1 May 1999

A four-sided biased die is used in a game. A player pays 10 counters to roll the die. The following table shows the possible scores on the die, the probability of each score and the number of counters the player receives in return for each score.

Score	1	2	3	4
Probability	$\frac{1}{2}$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{10}$
Number of counters the player receives	4	5	15	$n$

Find the value of $n$ so that the player has an expected return of 9 counters per roll.

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Answer: $n = 30$

Show working and answer

The expected return is the mean number of counters received: $E = 4 \cdot \frac{1}{2} + 5 \cdot \frac{1}{5} + 15 \cdot \frac{1}{5} + n \cdot \frac{1}{10} = 2 + 1 + 3 + \frac{n}{10} = 6 + \frac{n}{10} .$ Setting this equal to 9: $6 + \frac{n}{10} = 9 \Rightarrow n = 30.$

Answer: $n = 30$

Exercise 25 MediumIB · P1 TZ0 November 2013

The discrete random variable $X$ has the following probability distribution:

$x$	0	1	2	3
$P (X = x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$a$

a) Find the value of $a$ .

b) Find $E (X)$ .

c) Find $Var (X)$ .

Show answer

Answer: a) $a = \frac{1}{30}$ ; b) $E (X) = \frac{6}{5}$ ; c) $Var (X) = \frac{14}{25}$

Show working and answer

a) $\frac{1}{6} + \frac{1}{2} + \frac{3}{10} + a = 1$ , that is, $\frac{5 + 15 + 9}{30} + a = 1$ , so $a = \frac{1}{30}$ .

b) $E (X) = 0 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30} = \frac{1}{2} + \frac{3}{5} + \frac{1}{10} = \frac{6}{5} .$

c) $E (X^{2}) = \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30} = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = 2$ , so $Var (X) = 2 - (\frac{6}{5})^{2} = 2 - \frac{36}{25} = \frac{14}{25} .$

Answer: a) $a = \frac{1}{30}$ ; b) $E (X) = \frac{6}{5}$ ; c) $Var (X) = \frac{14}{25}$

Exercise 26 HardIB · P2 TZ0 November 2005

In a game a player pays an entry fee of $ $n$ . They then choose a number from 1, 2, 3, 4, 5, 6 and roll three standard dice. If the number they chose appears on all three dice, they win three times their entry fee. If the number appears on exactly two of the dice, they win twice their entry fee. If the number appears on exactly one of the dice, they win their entry fee. If the number does not appear on any of the dice, they win nothing.

a) Copy and complete the probability table below.

Gain ($)	$- n$	$n$	$2 n$	$3 n$
Probability		$\frac{75}{216}$

b) Show that the expected gain is $ $(- \frac{17 n}{216})$ .

c) What should the entry fee be so that the player's loss in each game is 34 cents?

Show answer

Answer: a) $\frac{125}{216}$ , $\frac{75}{216}$ , $\frac{15}{216}$ , $\frac{1}{216}$ ; b) $- \frac{17 n}{216}$ ; c) $n = 4.32$ $

Show working and answer

Let $G$ be the gain. Probabilities that the chosen number appears on $k$ of the three dice: $(\frac{1}{6})^{k} (\frac{5}{6})^{3 - k} (k 3)$ .

a) - None ( $- n$ ): $(\frac{5}{6})^{3} = \frac{125}{216}$ .

One ( $n$ ): $3 \cdot \frac{1}{6} \cdot (\frac{5}{6})^{2} = \frac{75}{216}$ (given).
Two ( $2 n$ ): $3 \cdot (\frac{1}{6})^{2} \cdot \frac{5}{6} = \frac{15}{216}$ .
Three ( $3 n$ ): $(\frac{1}{6})^{3} = \frac{1}{216}$ .

Gain ($)	$- n$	$n$	$2 n$	$3 n$
Probability	$\frac{125}{216}$	$\frac{75}{216}$	$\frac{15}{216}$	$\frac{1}{216}$

b) $E (G) = \frac{- 125 n + 75 n + 30 n + 3 n}{216} = \frac{- 17 n}{216} .$

c) The loss is 34 cents, i.e. $0.34$ dollars: $\frac{17 n}{216} = 0.34 \Rightarrow n = \frac{0.34 \cdot 216}{17} = 4.32.$

Answer: a) $\frac{125}{216}$ , $\frac{75}{216}$ , $\frac{15}{216}$ , $\frac{1}{216}$ ; b) $- \frac{17 n}{216}$ ; c) $n = 4.32$ $

Binomial distribution: probabilities

Exercise 27 BasicIB · P2 TZ2 May 2014

Six customers queue at a supermarket. Each customer can choose whether to pay in cash or by credit card. Assume that whether one customer pays by credit card or not is independent of the payment method chosen by the other customers. It is known that $60 %$ of customers choose to pay by credit card. Find the probability that:

(i) the first three customers pay by credit card and the next three pay in cash.

(ii) of the six customers, exactly three pay by credit card.

Show answer

Answer: i) $0.0138$ ; ii) $0.276$

Show working and answer

Let $T$ be the number of customers who pay by card among the 6: $T \sim B (6; 0.6)$ .

(i) It is one specific sequence (card, card, card, cash, cash, cash), with independence between customers: $0. 6^{3} \cdot 0. 4^{3} = 0.013824 \approx 0.0138.$

(ii) Exactly three by card, in any order: there are $(3 6) = 20$ sequences, each with probability $0. 6^{3} \cdot 0. 4^{3}$ : $P (T = 3) = (3 6) 0. 6^{3} 0. 4^{3} = 0.27648 \approx 0.276.$

Answer: i) $0.0138$ ; ii) $0.276$

Exercise 28 BasicIB · P1 May 1997

If 30% of the students at a college do not graduate, calculate the probability that, of 6 university students selected at random, exactly 4 of them graduate.

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Answer: $P (X = 4) \approx 0.324$

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If 30% do not graduate, then 70% do. Let $X$ be the number of graduates among the 6: $X \sim B (6; 0.7)$ . $P (X = 4) = (4 6) 0. 7^{4} 0. 3^{2} = 15 \cdot 0.2401 \cdot 0.09 = 0.324135 \approx 0.324.$

Answer: $P (X = 4) \approx 0.324$

Exercise 29 MediumIB · P1 TZ2 May 2010

An unbalanced coin is biased so that the probability of obtaining heads is $\frac{4}{7}$ . It is tossed 6 times, and $X$ represents the number of heads obtained. Find the value of the ratio $\frac{P ( X = 3 )}{P ( X = 2 )} .$

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Answer: $\frac{P ( X = 3 )}{P ( X = 2 )} = \frac{16}{9}$

Show working and answer

$X \sim B (6, \frac{4}{7})$ , so $P (X = k) = (k 6) (\frac{4}{7})^{k} (\frac{3}{7})^{6 - k}$ .

$\frac{P ( X = 3 )}{P ( X = 2 )} = \frac{( 3 6 ) ( \frac{4}{7} ) ^{3} ( \frac{3}{7} ) ^{3}}{( 2 6 ) ( \frac{4}{7} ) ^{2} ( \frac{3}{7} ) ^{4}} = \frac{20}{15} \cdot \frac{4/7}{3/7} = \frac{4}{3} \cdot \frac{4}{3} = \frac{16}{9} .$

Answer: $\frac{P ( X = 3 )}{P ( X = 2 )} = \frac{16}{9}$

Exercise 30 MediumIB · P2 TZ2 May 2008

Over a one-month period, Ava and Sven play a total of $n$ tennis matches. The probability that Ava wins a match is equal to $0.4$ . The result of each match played is independent of any other match played. Let $X$ be the number of matches won by Ava over a one-month period.

a) Find an expression for $P (X = 2)$ in terms of $n$ .

b) If the probability that Ava wins two matches is equal to $0.121$ , rounded to three decimal places, find the value of $n$ .

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Answer: a) $P (X = 2) = (2 n) 0. 4^{2} 0. 6^{n - 2}$ ; b) $n = 10$

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a) $X \sim B (n; 0.4)$ , so $P (X = 2) = (2 n) 0. 4^{2} 0. 6^{n - 2} .$

b) We must solve $(2 n) 0. 4^{2} 0. 6^{n - 2} = 0.121$ . Using the GDC (binomial distribution, point probability), values of $n$ are tested:

$n$	8	9	10	11
$P (X = 2)$	$0.209$	$0.161$	$0.121$	$0.089$

Only $n = 10$ gives $0.121$ (more precisely $0.12093$ ).

Answer: a) $P (X = 2) = (2 n) 0. 4^{2} 0. 6^{n - 2}$ ; b) $n = 10$

Exercise 31 MediumIB · P1 TZ2 May 2016

An unbalanced coin is tossed five times. On each toss, the probability of obtaining heads is equal to $p$ . Let $X$ be the number of heads obtained.

a) Find, in terms of $p$ , an expression for $P (X = 4)$ .

b) i) Determine the value of $p$ for which $P (X = 4)$ reaches a maximum value.

(ii) For this value of $p$ , determine the expected number of heads.

Show answer

Answer: a) $P (X = 4) = 5 p^{4} (1 - p)$ ; b) i) $p = \frac{4}{5}$ ; ii) $E (X) = 4$

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a) $X \sim B (5, p)$ : $P (X = 4) = (4 5) p^{4} (1 - p) = 5 p^{4} (1 - p) .$

b) (i) Let $f (p) = 5 p^{4} - 5 p^{5}$ . Differentiating: $f^{'} (p) = 20 p^{3} - 25 p^{4} = 5 p^{3} (4 - 5 p)$ . $f^{'} (p) = 0$ with $0 < p < 1$ gives $p = \frac{4}{5}$ . For $p < \frac{4}{5}$ we have $f^{'} > 0$ and for $p > \frac{4}{5}$ , $f^{'} < 0$ , so it is a maximum.

(ii) With $p = \frac{4}{5}$ : $E (X) = 5 \cdot \frac{4}{5} = 4$ .

Answer: a) $P (X = 4) = 5 p^{4} (1 - p)$ ; b) i) $p = \frac{4}{5}$ ; ii) $E (X) = 4$

Exercise 32 MediumIB · P1 May 2002

When Juan throws a stone at a target, the probability that he hits the target is $0.4$ . Juan throws a stone 6 times.

a) Find the probability that he hits the target exactly 4 times.

b) Find the probability that he hits the target for the first time on the third throw of a stone.

Show answer

Answer: a) $0.138$ ; b) $0.144$

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a) Let $X \sim B (6; 0.4)$ : $P (X = 4) = (4 6) 0. 4^{4} 0. 6^{2} = 15 \cdot 0.0256 \cdot 0.36 \approx 0.138.$

b) Two misses and then a hit: $P = 0.6 \cdot 0.6 \cdot 0.4 = 0.144.$

Answer: a) $0.138$ ; b) $0.144$

Exercise 33 MediumIB · P2 TZ1 May 2015

A mosaic is to be created by randomly selecting 1,000 small tiles, each of which is either black or white. The probability that a tile is white is $0.1$ . Let the random variable $W$ be the number of white tiles.

a) Calculate the distribution of $W$ , including the values of any parameters.

b) Write down the mean of $W$ .

c) Find $P (W > 89)$ .

Show answer

Answer: a) $W \sim B (1000; 0.1)$ , $P (W = x) = (x 1000) 0. 1^{x} 0. 9^{1000 - x}$ ; b) $E (W) = 100$ ; c) $P (W > 89) = 0.867$

Show working and answer

a) Each tile is white with probability $0.1$ , independently, and there are $n = 1000$ tiles: $W \sim B (1000; 0.1), P (W = x) = (x 1000) 0. 1^{x} 0. 9^{1000 - x}, x = 0, 1, \dots, 1000.$

b) $E (W) = n p = 1000 \cdot 0.1 = 100$ .

c) Using the GDC (binomial distribution): $P (W > 89) = 1 - P (W \leq 89) \approx 0.867.$

Answer: a) $W \sim B (1000; 0.1)$ , $P (W = x) = (x 1000) 0. 1^{x} 0. 9^{1000 - x}$ ; b) $E (W) = 100$ ; c) $P (W > 89) = 0.867$

Exercise 34 MediumIB · P1 November 2003

A TV channel broadcasts the news at the same time every day. The probability that Alicia watches the news on a given day is $0.4$ . Calculate the probability that, over five consecutive days, she watches the news on at most three days.

Show answer

Answer: $P (X \leq 3) = 0.913$

Show working and answer

Let $X \sim B (5; 0.4)$ be the number of days she watches the news. "At most three days" is the complement of "four or five days": $P (X \leq 3) = 1 - P (X = 4) - P (X = 5) = 1 - 5 \cdot 0. 4^{4} \cdot 0.6 - 0. 4^{5} .$ $= 1 - 0.0768 - 0.01024 = 0.91296 \approx 0.913.$

Answer: $P (X \leq 3) = 0.913$

Exercise 35 MediumIB · P1 May 2003

When a boy plays a funfair game, the probability that he wins a prize is $0.25$ . He plays the game 10 times. Let $X$ be the total number of prizes he wins. Assuming that the games are independent, find

a) $E (X)$ .

b) $P (X \leq 2)$ .

Show answer

Answer: a) $E (X) = 2.5$ ; b) $P (X \leq 2) = 0.526$

Show working and answer

$X \sim B (10; 0.25)$ .

a) $E (X) = n p = 10 \cdot 0.25 = 2.5$ .

b) $P (X \leq 2) = P (0) + P (1) + P (2)$ : $0.7 5^{10} + 10 \cdot 0.25 \cdot 0.7 5^{9} + 45 \cdot 0.2 5^{2} \cdot 0.7 5^{8} \approx 0.0563 + 0.1877 + 0.2816 \approx 0.526.$

Answer: a) $E (X) = 2.5$ ; b) $P (X \leq 2) = 0.526$

Exercise 36 MediumIB · P1 TZ1 May 2004

Marian fires ten arrows at a target. Each arrow has probability $0.4$ of hitting the target, independently of all the other arrows. Let $X$ be the number of these arrows that hit the target.

a) Find the mean and standard deviation of $X$ .

b) Calculate $P (X \geq 2)$ .

Show answer

Answer: a) mean $= 4$ , $σ = 2.4 \approx 1.55$ ; b) $P (X \geq 2) = 0.954$

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$X \sim B (10; 0.4)$ .

a) Mean: $n p = 4$ . Variance: $n pq = 10 \cdot 0.4 \cdot 0.6 = 2.4$ , so the standard deviation is $2.4 \approx 1.55$ .

b) By the complement: $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) = 1 - 0. 6^{10} - 10 \cdot 0.4 \cdot 0. 6^{9} \approx 1 - 0.00605 - 0.0403 \approx 0.954.$

Answer: a) mean $= 4$ , $σ = 2.4 \approx 1.55$ ; b) $P (X \geq 2) = 0.954$

Exercise 37 MediumIB · P1 TZ1 May 2005

In a 20-question multiple-choice exam, each question has four possible answers, only one of which is correct. Robert marks an answer to each question at random.

a) Find his expected number of correct answers.

b) Find the probability that Robert obtains this expected number of correct answers.

Show answer

Answer: a) $5$ ; b) $P (X = 5) = 0.202$

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$X \sim B (20; \frac{1}{4})$ is the number of correct answers.

a) $E (X) = n p = 20 \cdot \frac{1}{4} = 5$ .

b) $P (X = 5) = (5 20) (\frac{1}{4})^{5} (\frac{3}{4})^{15} = 15504 \cdot \frac{3 ^{15}}{4 ^{20}} \approx 0.202.$

Answer: a) $5$ ; b) $P (X = 5) = 0.202$

Exercise 38 MediumIB · P2 TZ0 November 2006

A bag contains a large number of coloured ribbons. One quarter of the ribbons are yellow, and the rest are blue. Ten ribbons are chosen at random from the bag.

a) Find the expected number of yellow ribbons chosen.

b) Find the probability that exactly six of these ribbons are yellow.

c) Find the probability that at least two of these ribbons are yellow.

d) Find the most probable number of yellow ribbons chosen.

e) What assumption have you made about the probability of choosing a yellow ribbon?

Show answer

Answer: a) $2.5$ ; b) $0.0162$ ; c) $0.756$ ; d) $2$ ; e) constant probability $\frac{1}{4}$ at each selection (independent selections)

Show working and answer

As there is a large number of ribbons, $X \sim B (10; \frac{1}{4})$ .

a) $E (X) = 10 \cdot \frac{1}{4} = 2.5$ .

b) Using the GDC (binomial distribution): $P (X = 6) = (6 10) (\frac{1}{4})^{6} (\frac{3}{4})^{4} \approx 0.0162.$

c) $P (X \geq 2) = 1 - P (X \leq 1) = 1 - 0.0563 - 0.1877 \approx 0.756.$

d) We look for the value of $x$ with the greatest probability (using the GDC): $P (0) = 0.0563$ , $P (1) = 0.188$ , $P (2) = 0.282$ , $P (3) = 0.250$ , $P (4) = 0.146$ , and it decreases from there. The most probable value is $x = 2$ .

e) It is assumed that the probability of choosing a yellow ribbon is always $\frac{1}{4}$ and that the selections are independent. This is reasonable because the bag contains many ribbons, so when ten are chosen the proportion of yellow ones barely changes.

Answer: a) $2.5$ ; b) $0.0162$ ; c) $0.756$ ; d) $2$ ; e) constant probability $\frac{1}{4}$ at each selection (independent selections)

Exercise 39 MediumIB · P1 TZ2 May 2007

A biology test consists of seven multiple-choice questions. Each question has five possible answers, of which only one is correct. To pass the test, a student must answer at least four questions correctly. John does not know the answers, so for each question he chooses an answer at random.

a) Find the probability that John answers exactly four questions correctly.

b) Find the probability that John passes the biology test.

Show answer

Answer: a) $0.0287$ ; b) $0.0333$

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Let $X \sim B (7; \frac{1}{5})$ be the number of correct answers when guessing at random. Passing means getting at least four correct.

a) $P (X = 4) = (4 7) (\frac{1}{5})^{4} (\frac{4}{5})^{3} = 35 \cdot \frac{64}{78125} = \frac{448}{15625} \approx 0.0287.$

b) $P (X \geq 4) = P (4) + P (5) + P (6) + P (7) .$ where $P (5) = 21 \cdot \frac{16}{78125}$ , $P (6) = 7 \cdot \frac{4}{78125}$ and $P (7) = \frac{1}{78125}$ : $P (X \geq 4) = \frac{2240 + 336 + 28 + 1}{78125} = \frac{2605}{78125} = \frac{521}{15625} \approx 0.0333.$

Answer: a) $0.0287$ ; b) $0.0333$

Exercise 40 MediumIB · P2 TZ0 November 2011

The probability that the 08:00 train arrives late on a weekday (Monday to Friday) is equal to $\frac{1}{10}$ . Assuming that delays occur independently of one another,

a) find the probability that, over a period of five weekdays, the 08:00 train arrives late exactly twice.

b) find the minimum number of weekdays for which the probability that the 08:00 train arrives late at least once is greater than $90 %$ .

Show answer

Answer: a) $0.0729$ ; b) $n = 22$ days

Show working and answer

a) $X \sim B (5; 0.1)$ : $P (X = 2) = (2 5) 0. 1^{2} 0. 9^{3} = 10 \cdot 0.01 \cdot 0.729 = 0.0729.$

b) With $n$ weekdays, $P (at least one delay) = 1 - 0. 9^{n} > 0.9$ , that is, $0. 9^{n} < 0.1$ . Using the GDC (or taking logarithms): $n > \frac{ln 0.1}{ln 0.9} \approx 21.85.$ Check: $0. 9^{21} \approx 0.109 > 0.1$ and $0. 9^{22} \approx 0.098 < 0.1$ . The minimum is $n = 22$ .

Answer: a) $0.0729$ ; b) $n = 22$ days

Exercise 41 MediumIB · P2 TZ0 May 2006

Andrew fires 20 arrows at a target. He has a probability of hitting the target of $0.3$ . All shots are independent of each other. Let $X$ be the number of arrows that hit the target.

a) Find the mean and standard deviation of $X$ .

b) Find

(i) $P (X = 5)$ .

(ii) $P (4 \leq X \leq 8)$ .

Bill also fires arrows at a target, with a probability of hitting the target of $0.3$ . All shots are independent.

c) Calculate the probability that Bill hits the target for the first time on his third shot.

d) Calculate the minimum number of shots required so that the probability that at least one shot hits the target is greater than $0.99$ .

Show answer

Answer: a) mean $= 6$ , $σ \approx 2.05$ ; b) i) $0.179$ ; ii) $0.780$ ; c) $0.147$ ; d) $n = 13$

Show working and answer

$X \sim B (20; 0.3)$ .

a) Mean: $n p = 6$ . Standard deviation: $n pq = 20 \cdot 0.3 \cdot 0.7 = 4.2 \approx 2.05$ .

b) Using the GDC (binomial distribution):

(i) $P (X = 5) = (5 20) 0. 3^{5} 0. 7^{15} \approx 0.179$ .

(ii) $P (4 \leq X \leq 8) = P (X \leq 8) - P (X \leq 3) \approx 0.780$ .

c) Two misses and a hit: $0.7 \cdot 0.7 \cdot 0.3 = 0.147$ .

d) With $n$ shots, $P (at least one hit) = 1 - 0. 7^{n} > 0.99$ , that is, $0. 7^{n} < 0.01$ . Using the GDC: $0. 7^{12} \approx 0.0138 > 0.01$ and $0. 7^{13} \approx 0.0097 < 0.01$ . The minimum is $n = 13$ shots.

Answer: a) mean $= 6$ , $σ \approx 2.05$ ; b) i) $0.179$ ; ii) $0.780$ ; c) $0.147$ ; d) $n = 13$

Exercise 42 MediumIB · P2 TZ0 November 2021

A factory makes lamps. It is known that the probability that a lamp is defective is $0.05$ . A random sample of 30 lamps is chosen.

a) Find the probability that there is at least one defective lamp in the sample.

b) Given that there is at least one defective lamp in the sample, find the probability that there are at most two defective lamps.

Show answer

Answer: a) $0.785$ ; b) $0.761$

Show working and answer

$X \sim B (30; 0.05)$ is the number of defective lamps.

a) $P (X \geq 1) = 1 - P (X = 0) = 1 - 0.9 5^{30} \approx 0.785.$

b) Conditional probability: $P (X \leq 2 ∣ X \geq 1) = \frac{P ( X = 1 ) + P ( X = 2 )}{P ( X \geq 1 )} .$ Using the GDC (binomial distribution): $P (X = 1) \approx 0.3389$ and $P (X = 2) \approx 0.2586$ , so $P = \frac{0.3389 + 0.2586}{0.7854} \approx 0.761.$

Answer: a) $0.785$ ; b) $0.761$

Exercise 43 HardIB · P1 May 1996

How many times must a pair of dice be rolled so that the probability of obtaining a double, that is, the same number on both dice, is greatest?

Show answer

Answer: 5 or 6 rolls (both give the maximum probability, $\approx 0.402$ )

Show working and answer

We interpret this as: how many times $n$ must the pair of dice be rolled so that the probability of obtaining exactly one double is maximal?

In one roll, $P (double) = \frac{6}{36} = \frac{1}{6}$ . Let $X$ be the number of doubles in $n$ rolls: $X \sim B (n, \frac{1}{6})$ and $P_{n} = P (X = 1) = n \cdot \frac{1}{6} (\frac{5}{6})^{n - 1} .$

Each value is compared with the next one: $\frac{P _{n + 1}}{P _{n}} = \frac{n + 1}{n} \cdot \frac{5}{6} .$ This ratio is greater than 1 if $n < 5$ , equal to 1 if $n = 5$ and less than 1 if $n > 5$ . Thus $P_{n}$ increases up to $n = 5$ , $P_{5} = P_{6}$ , and then decreases.

Check: $P_{5} = 5 \cdot \frac{1}{6} (\frac{5}{6})^{4} = \frac{3125}{7776} \approx 0.402 = P_{6}$ , greater than $P_{4} \approx 0.386$ and than $P_{7} \approx 0.391$ .

Answer: 5 or 6 rolls (both give the maximum probability, $\approx 0.402$ )

Exercise 44 HardIB · P2 November 1996

(i) An experiment has two possible outcomes: a success occurs with probability $p$ , $(0 < p < 1)$ , and a failure occurs with probability $1 - p$ . The outcomes of successive performances of this experiment are independent.

a) Find the value of $p$ if the probability of obtaining exactly two successes in six experiments is three times greater than the probability of obtaining exactly three successes in six experiments.

b) In another similar experiment, the value of $p$ was found to be $\frac{1}{5}$ . Find the smallest number of times the experiment must be performed so that the probability of obtaining at least one success is greater than $0.99$ .

(ii) Six per cent ( $6%$ ) of the light bulbs produced in a given factory are defective. The manufacturer chooses 200 bulbs from one day's production.

a) If $X$ represents the number of defective bulbs found in the sample, find the mean and standard deviation of $X$ .

b) Determine the probability that 20 or more defective bulbs are found in the sample of 200 bulbs.

Show answer

Answer: i) a) $p = \frac{1}{5}$ ; b) $n = 21$ ; ii) a) $E (X) = 12$ , $σ \approx 3.36$ ; b) $P (X \geq 20) \approx 0.0179$

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(i) a) Let $X \sim B (6, p)$ . The condition is $P (X = 2) = 3 P (X = 3)$ : $(2 6) p^{2} (1 - p)^{4} = 3 (3 6) p^{3} (1 - p)^{3} \Rightarrow 15 (1 - p) = 60 p \Rightarrow 1 - p = 4 p \Rightarrow p = \frac{1}{5} .$

b) With $p = \frac{1}{5}$ and $n$ repetitions, $P (at least one success) = 1 - 0. 8^{n} > 0.99$ , that is, $0. 8^{n} < 0.01$ : $n > \frac{ln 0.01}{ln 0.8} \approx 20.6.$ Check: $1 - 0. 8^{20} = 0.9885 < 0.99$ and $1 - 0. 8^{21} = 0.9908 > 0.99$ . The smallest value is $n = 21$ .

(ii) a) $X \sim B (200; 0.06)$ : $E (X) = 200 \cdot 0.06 = 12$ and $σ = 200 \cdot 0.06 \cdot 0.94 = 11.28 \approx 3.36$ .

b) $P (X \geq 20) = 1 - P (X \leq 19)$ . Using the GDC (binomial distribution, cumulative probability): $P (X \leq 19) \approx 0.9821$ , so $P (X \geq 20) \approx 0.0179 \approx 0.018$ .

Answer: i) a) $p = \frac{1}{5}$ ; b) $n = 21$ ; ii) a) $E (X) = 12$ , $σ \approx 3.36$ ; b) $P (X \geq 20) \approx 0.0179$

Exercise 45 HardIB · P1 TZ0 November 2017

Chloe and Selena play a game in which each has four cards showing the capital letters A, B, C and D. Chloe places her cards face up on the table in the order A, B, C and D, as shown in the diagram.

Two rows of four cards aligned: Chloe's row, with cards A, B, C and D face up from left to right, and below it Selena's row, with four identical cards face down (striped)

Selena shuffles her cards and places them face down on the table. Then she turns them over one by one to see whether her card matches the card of Chloe directly above it.

Chloe wins if there are no matches; otherwise, Selena wins.

a) Show that the probability that Chloe wins the game is $\frac{3}{8}$ .

Chloe and Selena repeat their game so that they play a total of 50 times. Let $X$ be the discrete random variable representing the number of times Chloe wins.

b) Determine

(i) the mean of $X$ ;

(ii) the variance of $X$ .

Show answer

Answer: a) $\frac{3}{8}$ ; b) i) $E (X) = \frac{75}{4}$ ; ii) $Var (X) = \frac{375}{32}$

Show working and answer

a) Selena's cards can be arranged in $4! = 24$ equally likely ways. Chloe wins when no card matches the one above it (arrangements with no fixed points). By inclusion–exclusion: $4! - (1 4) 3! + (2 4) 2! - (3 4) 1! + (4 4) 0! = 24 - 24 + 12 - 4 + 1 = 9.$ Hence $P (Chloe wins) = \frac{9}{24} = \frac{3}{8}$ .

b) $X \sim B (50, \frac{3}{8})$ .

(i) $E (X) = 50 \cdot \frac{3}{8} = \frac{75}{4} = 18.75$ .

(ii) $Var (X) = 50 \cdot \frac{3}{8} \cdot \frac{5}{8} = \frac{750}{64} = \frac{375}{32} \approx 11.7$ .

Answer: a) $\frac{3}{8}$ ; b) i) $E (X) = \frac{75}{4}$ ; ii) $Var (X) = \frac{375}{32}$

Exercise 46 HardIB · P1 May 2001

$X$ is a binomial random variable where the number of trials is 5 and the probability of success on each trial is $p$ . Calculate the values of $p$ if $P (X = 4) = 0.12$ .

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Answer: $p \approx 0.459$ or $p \approx 0.973$

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For $X \sim B (5, p)$ : $P (X = 4) = 5 p^{4} (1 - p)$ . We must solve $5 p^{4} (1 - p) = 0.12, 0 < p < 1.$ Using the GDC, plot $y = 5 x^{4} (1 - x)$ and the line $y = 0.12$ , and find the intersections (the curve rises up to $x = 0.8$ and falls afterwards, so there are two): $p \approx 0.459 or p \approx 0.973.$

Answer: $p \approx 0.459$ or $p \approx 0.973$

Exercise 47 HardIB · P2 TZ1 May 2016

Six balls numbered 1, 2, 2, 3, 3, 3 are placed in a bag. The balls are taken out one at a time at random and the number is recorded. Throughout the question, a ball is always replaced before the next ball is taken.

a) One ball is taken from the bag. Let $X$ be the value shown on the ball. Calculate $E (X)$ .

b) Three balls are taken from the bag. Find the probability that

(i) the sum of the three numbers is 5;

(ii) the median of the three numbers is 1.

c) Ten balls are taken from the bag. Find the probability that fewer than four of the balls are numbered 2.

d) Find the smallest number of balls that must be taken from the bag so that the probability of obtaining at least one ball numbered 2 is greater than $0.95$ .

e) Another bag also contains balls numbered 1, 2 or 3. Eight balls are taken from this bag at random. It is calculated that the expected number of balls numbered 1 is $4.8$ , and the variance of the number of balls numbered 2 is $1.5$ . Calculate the smallest possible number of balls numbered 3 in this bag.

Show answer

Answer: a) $E (X) = \frac{7}{3}$ ; b) i) $\frac{7}{72}$ ; ii) $\frac{2}{27}$ ; c) $0.559$ ; d) $n = 8$ ; e) $3$ balls

Show working and answer

Probability of each number: $P (1) = \frac{1}{6}$ , $P (2) = \frac{2}{6} = \frac{1}{3}$ , $P (3) = \frac{3}{6} = \frac{1}{2}$ .

a) $E (X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{2} = \frac{1 + 4 + 9}{6} = \frac{14}{6} = \frac{7}{3}$ .

b) (i) A sum of 5 with three numbers from ${1, 2, 3}$ : the combinations are ${1, 1, 3}$ and ${1, 2, 2}$ .

${1, 1, 3}$ : $3 \cdot (\frac{1}{6})^{2} \cdot \frac{1}{2} = \frac{3}{72}$ .
${1, 2, 2}$ : $3 \cdot \frac{1}{6} \cdot (\frac{1}{3})^{2} = \frac{3}{54} = \frac{4}{72}$ .

$P = \frac{3}{72} + \frac{4}{72} = \frac{7}{72} .$

(ii) The median of three numbers is 1 if at least two of them are 1. With $Z$ = number of 1s in three draws, $Z \sim B (3, \frac{1}{6})$ : $P (Z \geq 2) = 3 (\frac{1}{6})^{2} \frac{5}{6} + (\frac{1}{6})^{3} = \frac{15}{216} + \frac{1}{216} = \frac{16}{216} = \frac{2}{27} .$

c) Let $W$ be the number of balls with a 2 in ten draws: $W \sim B (10, \frac{1}{3})$ . Using the GDC (binomial distribution, cumulative probability): $P (W < 4) = P (W \leq 3) \approx 0.559.$

d) With $n$ draws: $P (at least one 2) = 1 - (\frac{2}{3})^{n} > 0.95$ . Using the GDC, we compare:

$n$	7	8
$1 - (\frac{2}{3})^{n}$	$0.941$	$0.961$

The smallest value is $n = 8$ .

e) Let $p_{1}, p_{2}, p_{3}$ be the proportions of balls numbered 1, 2 and 3 in the second bag; 8 balls are taken.

Expected number of 1s: $8 p_{1} = 4.8 \Rightarrow p_{1} = 0.6$ .
Variance of 2s: $8 p_{2} (1 - p_{2}) = 1.5 \Rightarrow p_{2} (1 - p_{2}) = 0.1875 \Rightarrow p_{2} = 0.25$ or $p_{2} = 0.75$ . Since $p_{1} + p_{2} \leq 1$ and $p_{1} = 0.6$ , only $p_{2} = 0.25$ is valid.
Hence $p_{3} = 1 - 0.6 - 0.25 = 0.15$ .

The proportions $0.6; 0.25; 0.15$ are $\frac{12}{20}, \frac{5}{20}, \frac{3}{20}$ . The total number of balls must be a multiple of 20, and the smallest is 20, with $3$ balls numbered 3.

Answer: a) $E (X) = \frac{7}{3}$ ; b) i) $\frac{7}{72}$ ; ii) $\frac{2}{27}$ ; c) $0.559$ ; d) $n = 8$ ; e) $3$ balls

Exercise 48 HardIB · P2 TZ1 May 2014

a) Find the term in $x^{5}$ in the expansion of $(3 x + A) (2 x + B)^{6}$ .

Mina and Norbert each have a fair six-sided die with the faces labelled 1, 2, 3, 4, 5 and 6; they roll to decide whether they will eat a biscuit. Mina rolls it only once and eats a biscuit if she gets a four, five or six. Norbert rolls his die six times and each time he eats a biscuit if he gets a five or a six.

b) Calculate the probability that five biscuits are eaten.

Show answer

Answer: a) $720 B^{2} + 192 A B$ ; b) $\frac{4}{81} \approx 0.0494$

Show working and answer

a) The term in $x^{5}$ comes from two products:

$3 x$ times the term in $x^{4}$ of $(2 x + B)^{6}$ : $(4 6) (2 x)^{4} B^{2} = 15 \cdot 16 B^{2} x^{4} = 240 B^{2} x^{4}$ , giving $720 B^{2} x^{5}$ .
$A$ times the term in $x^{5}$ : $(5 6) (2 x)^{5} B = 6 \cdot 32 B x^{5} = 192 B x^{5}$ , giving $192 A B x^{5}$ .

The term is $(720 B^{2} + 192 A B) x^{5}$ .

b) Mina eats a biscuit with probability $\frac{3}{6} = \frac{1}{2}$ (0 or 1 biscuit). Norbert eats one with probability $\frac{2}{6} = \frac{1}{3}$ on each roll: $N \sim B (6; \frac{1}{3})$ . There are five biscuits if:

Mina eats one and Norbert eats 4, or
Mina does not eat one and Norbert eats 5.

Using the GDC (binomial distribution): $P (N = 4) = 15 \cdot \frac{1}{81} \cdot \frac{4}{9} = \frac{60}{729}, P (N = 5) = 6 \cdot \frac{1}{243} \cdot \frac{2}{3} = \frac{12}{729} .$ $P = \frac{1}{2} \cdot \frac{60}{729} + \frac{1}{2} \cdot \frac{12}{729} = \frac{36}{729} = \frac{4}{81} \approx 0.0494.$

Answer: a) $720 B^{2} + 192 A B$ ; b) $\frac{4}{81} \approx 0.0494$

Exercise 49 HardIB · P1 TZ1 May 2013

On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any of these games is $\frac{2}{3}$ .

a) Show that the probability that Alfred wins exactly 4 of the games is $\frac{80}{243}$ .

b) (i) Explain why the total number of possible outcomes for the results of the 6 games is 64.

(ii) By expanding $(1 + x)^{6}$ and choosing a suitable value for $x$ , prove that $64 = (0 6) + (1 6) + (2 6) + (3 6) + (4 6) + (5 6) + (6 6) .$

(iii) State the meaning of this equality in the context of the 6 games played.

c) The next day, Alfred and Beatrice play the 6 games again. Assume that the probability that Alfred wins any of these games is still $\frac{2}{3}$ .

(i) Find an expression for the probability that Alfred wins 4 games on the first day and 2 on the second day. Give your answer in the form $(r 6)^{2} (\frac{2}{3})^{s} (\frac{1}{3})^{t}$ , where $r$ , $s$ and $t$ are to be found.

(ii) Using your answer to (c)(i) and 6 similar expressions, calculate the probability that Alfred wins a total of 6 games over the two days, as the sum of 7 probabilities.

(iii) Using the above, prove that $(6 12) = (0 6)^{2} + (1 6)^{2} + (2 6)^{2} + (3 6)^{2} + (4 6)^{2} + (5 6)^{2} + (6 6)^{2} .$

d) Alfred and Beatrice play $n$ games. Let $A$ be the number of games that Alfred wins. The expected value of $A$ can be written as $E (A) = r = 0 \sum n r (r n) \frac{a ^{r}}{b ^{n}} .$

(i) Find the values of a and b.

(ii) By differentiating the expansion of $(1 + x)^{n}$ , prove that the expected number of games won by Alfred is $\frac{2 n}{3}$ .

Show answer

Answer: a) $\frac{80}{243}$ ; b) (i) $2^{6} = 64$ outcomes; (iii) $(k 6)$ counts the outcomes with exactly $k$ wins for Alfred; c) (i) $r = 2$ , $s = 6$ , $t = 6$ ; (ii) $\frac{19712}{177147}$ ; (iii) shown; d) (i) $a = 2$ , $b = 3$ ; (ii) $E (A) = \frac{2 n}{3}$

Show working and answer

a) $A \sim B (6; \frac{2}{3})$ : $P (A = 4) = (4 6) (\frac{2}{3})^{4} (\frac{1}{3})^{2} = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729} = \frac{80}{243} .$

b) (i) Each game has 2 possible outcomes (Alfred wins or Beatrice wins), so the 6 games have $2^{6} = 64$ possible outcomes.

(ii) Setting $x = 1$ in the binomial expansion $(1 + x)^{6} = (0 6) + (1 6) x + \dots + (6 6) x^{6}$ gives $2^{6} = 64 = (0 6) + (1 6) + (2 6) + (3 6) + (4 6) + (5 6) + (6 6)$ .

(iii) The term $(k 6)$ is the number of outcomes in which Alfred wins exactly $k$ of the 6 games; the sum of all of them gives the total number of possible outcomes, 64.

c) (i) Alfred wins 4 on the first day and 2 on the second, independently: $(4 6) (\frac{2}{3})^{4} (\frac{1}{3})^{2} \cdot (2 6) (\frac{2}{3})^{2} (\frac{1}{3})^{4} = (2 6)^{2} (\frac{2}{3})^{6} (\frac{1}{3})^{6},$ using $(4 6) = (2 6)$ . That is, $r = 2$ , $s = 6$ , $t = 6$ .

(ii) The other ways of winning 6 games in total are $(k, 6 - k)$ for $k = 0, \dots, 6$ , each with probability $(k 6) (6 - k 6) (\frac{2}{3})^{6} (\frac{1}{3})^{6} = (k 6)^{2} (\frac{2}{3})^{6} (\frac{1}{3})^{6}$ . Summing the 7 probabilities: $P = [(0 6)^{2} + (1 6)^{2} + \dots + (6 6)^{2}] (\frac{2}{3})^{6} (\frac{1}{3})^{6} .$ Calculating, the sum in brackets is $1 + 36 + 225 + 400 + 225 + 36 + 1 = 924$ and $P = 924 \cdot \frac{64}{531441} = \frac{19712}{177147} .$

(iii) Over the two days Alfred plays 12 games, each with probability $\frac{2}{3}$ , so the number of wins is $B (12; \frac{2}{3})$ . Then $P (6 wins) = (6 12) (\frac{2}{3})^{6} (\frac{1}{3})^{6} .$ Equating this with the result of (ii) and dividing by $(\frac{2}{3})^{6} (\frac{1}{3})^{6}$ : $(6 12) = (0 6)^{2} + (1 6)^{2} + (2 6)^{2} + (3 6)^{2} + (4 6)^{2} + (5 6)^{2} + (6 6)^{2} .$

d) (i) $P (A = r) = (r n) (\frac{2}{3})^{r} (\frac{1}{3})^{n - r} = (r n) \frac{2 ^{r}}{3 ^{n}}$ , so $a = 2$ and $b = 3$ .

(ii) Differentiating $(1 + x)^{n} = \sum_{r = 0}^{n} (r n) x^{r}$ : $n (1 + x)^{n - 1} = r = 1 \sum n r (r n) x^{r - 1} .$ Multiplying by $x$ and taking $x = 2$ : $2 n \cdot 3^{n - 1} = r = 0 \sum n r (r n) 2^{r} .$ Dividing by $3^{n}$ : $E (A) = r = 0 \sum n r (r n) \frac{2 ^{r}}{3 ^{n}} = \frac{2 n \cdot 3 ^{n - 1}}{3 ^{n}} = \frac{2 n}{3} .$

Binomial: expected value, variance and finding n or p

Exercise 50 BasicIB · P1 November 2001

A coin is biased so that the probability of obtaining heads is $\frac{2}{3}$ . The coin is tossed 1,800 times. Let $X$ be the number of heads obtained. Calculate

a) the mean of $X$ ;

b) the standard deviation of $X$ .

Show answer

Answer: a) $E (X) = 1200$ ; b) $σ = 20$

Show working and answer

$X \sim B (1800, \frac{2}{3})$ .

a) $E (X) = n p = 1800 \cdot \frac{2}{3} = 1200$ .

b) $Var (X) = n p (1 - p) = 1800 \cdot \frac{2}{3} \cdot \frac{1}{3} = 400$ , so $σ = 400 = 20$ .

Answer: a) $E (X) = 1200$ ; b) $σ = 20$

Exercise 51 MediumIB · P2 TZ2 May 2013

Let the random variable $X \sim B (n, p)$ have mean 4 and variance 3.

(i) Determine $n$ and $p$ .

(ii) Find the probability that, in a single experiment, the result is 1 or 3.

Show answer

Answer: i) $n = 16$ , $p = \frac{1}{4}$ ; ii) $P (X = 1) + P (X = 3) \approx 0.261$

Show working and answer

(i) For $X \sim B (n, p)$ : mean $n p = 4$ and variance $n p (1 - p) = 3$ . Dividing, $1 - p = \frac{3}{4} \Rightarrow p = \frac{1}{4}, n = \frac{4}{p} = 16.$

(ii) With $X \sim B (16, \frac{1}{4})$ , “the result is 1 or 3” means $P (X = 1) + P (X = 3)$ : $P (X = 1) = (1 16) \frac{1}{4} (\frac{3}{4})^{15} \approx 0.0535, P (X = 3) = (3 16) (\frac{1}{4})^{3} (\frac{3}{4})^{13} \approx 0.2079.$ $P (X = 1) + P (X = 3) \approx 0.261.$

Answer: i) $n = 16$ , $p = \frac{1}{4}$ ; ii) $P (X = 1) + P (X = 3) \approx 0.261$

Exercise 52 MediumIB · P1 TZ1 May 2007

In an experiment, a trial is repeated $n$ times. The trials are independent and the probability $p$ of success on each trial is constant. Let $X$ be the number of successes in the $n$ trials. The mean of $X$ is $0.4$ and the standard deviation is $0.6$ .

a) Calculate $p$ .

b) Calculate $n$ .

Show answer

Answer: a) $p = \frac{1}{10}$ ; b) $n = 4$

Show working and answer

For a binomial distribution, $E (X) = n p = 0.4$ and $σ^{2} = n pq = 0. 6^{2} = 0.36$ .

a) Dividing: $q = \frac{n pq}{n p} = \frac{0.36}{0.4} = 0.9$ , so $p = 1 - q = 0.1$ .

b) From $n p = 0.4$ with $p = 0.1$ : $n = \frac{0.4}{0.1} = 4$ .

Answer: a) $p = \frac{1}{10}$ ; b) $n = 4$

Exercise 53 HardIB · P2 TZ2 May 2018

The random variable $X$ follows a binomial distribution with parameters $n$ and $p$ . It is known that $E (X) = 3.5$ .

a) Find the smallest possible value of $n$ .

It is also known that $P (X \leq 1) = 0.09478$ correct to 4 significant figures.

b) Determine the value of $n$ and the value of $p$ .

Show answer

Answer: a) $n = 4$ ; b) $n = 12$ , $p = \frac{7}{24} \approx 0.292$

Show working and answer

a) For a binomial distribution, $E (X) = n p = 3.5$ , so $p = \frac{3.5}{n}$ . Since $p \leq 1$ , we need $n \geq 3.5$ . The smallest integer is $n = 4$ (with $p = 0.875$ ).

b) We test values of $n$ with $p = \frac{3.5}{n}$ and calculate $P (X \leq 1)$ using the GDC (binomial distribution, cumulative probability):

$n$	10	11	12	13
$P (X \leq 1)$	$0.0860$	$0.0908$	$0.09478$	$0.0981$

The probability increases with $n$ and coincides with $0.09478$ for $n = 12$ , with $p = \frac{3.5}{12} = \frac{7}{24} \approx 0.292$ .

Answer: a) $n = 4$ ; b) $n = 12$ , $p = \frac{7}{24} \approx 0.292$

Combined contextual problems

Exercise 54 HardIB · P2 TZ1 May 2019

A café serves sandwiches and cakes. Each customer will choose one of the following three options: buy only a sandwich, buy only a cake, or buy both a sandwich and a cake. The probability that a customer buys a sandwich is $0.72$ and the probability that a customer buys a cake is $0.45$ .

a) Determine the probability that a customer chosen at random buys

(i) a sandwich and a cake;

(ii) only a sandwich.

On a normal day, 200 customers come to the café.

b) Find

(i) the expected number of cakes sold on a normal day;

(ii) the probability that more than 100 cakes are sold on a normal day.

It is known that 46% of the customers who visit the café are men, and that 80% of these men buy a sandwich.

c) (i) A customer is chosen at random. Find the probability that the customer is a man and buys a sandwich.

(ii) A female customer is chosen at random. Find the probability that she buys a sandwich.

Show answer

Answer: a) i) $0.17$ ; ii) $0.55$ ; b) i) $90$ ; ii) $0.0681$ ; c) i) $0.368$ ; ii) $0.652$

Show working and answer

Every customer buys at least one thing, so $P (S \cup C) = 1$ (S = sandwich, C = cake).

a) (i) $P (S \cup C) = P (S) + P (C) - P (S \cap C)$ , that is, $1 = 0.72 + 0.45 - P (S \cap C)$ . Thus $P (S \cap C) = 0.17.$

(ii) Only a sandwich: $P (S) - P (S \cap C) = 0.72 - 0.17 = 0.55$ .

b) Let $Y$ be the number of cakes sold among 200 customers: $Y \sim B (200; 0.45)$ .

(i) $E (Y) = 200 \cdot 0.45 = 90$ .

(ii) $P (Y > 100) = P (Y \geq 101)$ . Using the GDC (binomial distribution, cumulative probability), $P (Y \leq 100) \approx 0.9319$ , so $P (Y > 100) \approx 0.0681.$

c) (i) $P (man and sandwich) = 0.46 \cdot 0.8 = 0.368$ .

(ii) Women: $1 - 0.46 = 0.54$ . In total $0.72$ buy a sandwich, of which $0.368$ are men, so $0.72 - 0.368 = 0.352$ are women. $P (S ∣ woman) = \frac{0.352}{0.54} \approx 0.652.$

Answer: a) i) $0.17$ ; ii) $0.55$ ; b) i) $90$ ; ii) $0.0681$ ; c) i) $0.368$ ; ii) $0.652$

Exercise 55 HardIB · P2 TZ1 May 2018

Each of the $25$ students in a class is asked how many pets they have. Two students have three pets and no student has more than three pets. The mean and the standard deviation of the number of pets owned by the students in the class are $\frac{18}{25}$ and $\frac{24}{25}$ respectively. Find the number of students in the class who do not have a pet.

Show answer

Answer: $14$ students without a pet

Show working and answer

Let $n_{0}, n_{1}, n_{2}$ be the numbers of students with 0, 1 and 2 pets; there are $n_{3} = 2$ with 3 pets.

Total: $n_{0} + n_{1} + n_{2} + 2 = 25$ .
Mean: $\frac{n _{1} + 2 n _{2} + 3 \cdot 2}{25} = \frac{18}{25}$ , that is, $n_{1} + 2 n_{2} = 12$ .
Standard deviation: $Var = (\frac{24}{25})^{2} = \frac{576}{625}$ and $E (X^{2}) = Var + μ^{2} = \frac{576 + 324}{625} = \frac{36}{25}$ . Then $\frac{n _{1} + 4 n _{2} + 9 \cdot 2}{25} = \frac{36}{25}$ , that is, $n_{1} + 4 n_{2} = 18$ .

Subtracting the last two equations: $2 n_{2} = 6$ , so $n_{2} = 3$ and $n_{1} = 6$ .

Then $n_{0} = 25 - 6 - 3 - 2 = 14$ .

Answer: $14$ students without a pet

Exercise 56 HardIB · P1 May 1998

A student travels to school by bus, train or taxi. They roll a six-sided die to decide which method of transport to use. If they obtain a 1, 2 or 3 on the die, they travel by bus and the probability of arriving late to school is $10 %$ . If they obtain a 4 or 5 on the die, they travel by train and the probability of arriving late is $5 %$ . If they obtain a 6, they travel by taxi and the probability of arriving late is $2 %$ . The cost of each method of transport for a trip to school is: bus $0.50; train $1.80; and taxi $9.00. This information is shown in the table below.

Dice number	Method of transport	Probability of arriving late	Cost
1, 2 or 3	Bus	$10 %$	$0.50
4 or 5	Train	$5 %$	$1.80
6	Taxi	$2 %$	$9.00

a) Calculate the probability that the student arrives late to school on a randomly chosen day.

b) On a randomly chosen day, it is known that the student arrived late to school. Find the probability that they travelled to school by bus on that day.

c) In one year, the student makes 180 trips to school. Find the expected total cost of these 180 trips.

Show answer

Answer: a) $0.07$ ; b) $\frac{5}{7} \approx 0.714$ ; c) 423 $

Show working and answer

Probabilities of each method of transport: bus $\frac{3}{6} = \frac{1}{2}$ , train $\frac{2}{6} = \frac{1}{3}$ , taxi $\frac{1}{6}$ .

a) By the law of total probability: $P (late) = \frac{1}{2} \cdot 0.10 + \frac{1}{3} \cdot 0.05 + \frac{1}{6} \cdot 0.02 = 0.05 + 0.01 6 + 0.00 3 = 0.07.$

b) By Bayes' theorem: $P (bus ∣ late) = \frac{\frac{1}{2} \cdot 0.10}{0.07} = \frac{0.05}{0.07} = \frac{5}{7} \approx 0.714.$

c) Expected cost of one trip: $\frac{1}{2} \cdot 0.50 + \frac{1}{3} \cdot 1.80 + \frac{1}{6} \cdot 9.00 = 0.25 + 0.60 + 1.50 = 2.35 $.$ For 180 trips: $180 \cdot 2.35 = 423 $$ .

Answer: a) $0.07$ ; b) $\frac{5}{7} \approx 0.714$ ; c) 423 $

Exercise 57 HardIB · P2 November 1998

(i) A pack of twelve light bulbs contains three defective light bulbs. Two light bulbs are selected at random from the pack.

a) Show that the probability that none of these light bulbs is defective is $\frac{6}{11}$ .

b) Calculate the probability that exactly one of the first two light bulbs selected is defective.

The pack is accepted or rejected according to the following process. If none of the first two light bulbs selected is defective, the pack is accepted. If the first two light bulbs selected are both defective, the pack is rejected. If exactly one of the first two light bulbs selected is defective, a third light bulb is selected from the pack. If this light bulb is found to be defective, the pack is rejected; otherwise, the pack is accepted.

c) Given that exactly one of the first two light bulbs selected is defective, calculate the probability that the third light bulb selected is not defective.

d) Calculate the probability that the pack is finally accepted.

e) If the pack is finally accepted, calculate the probability that one of the first two light bulbs selected was defective.

(ii) A discrete probability distribution is defined by $P (X = x) = k x$ , where $x = 1, 2, 3, \dots, n$ and $k$ is a constant.

a) Show that $k = \frac{2}{n ( n + 1 )}$ .

b) Using the result $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ , calculate the mean of the distribution in terms of $n$ .

Show answer

Answer: i) a) $\frac{6}{11}$ ; b) $\frac{9}{22}$ ; c) $\frac{4}{5}$ ; d) $\frac{48}{55}$ ; e) $\frac{3}{8}$ ; ii) a) $k = \frac{2}{n ( n + 1 )}$ ; b) $E (X) = \frac{2 n + 1}{3}$

Show working and answer

(i) There are 12 light bulbs: 3 defective and 9 non-defective.

a) Without replacement, both are non-defective: $\frac{9}{12} \cdot \frac{8}{11} = \frac{72}{132} = \frac{6}{11} .$

b) Exactly one defective: (defective, non-defective) or (non-defective, defective): $2 \cdot \frac{3}{12} \cdot \frac{9}{11} = \frac{54}{132} = \frac{9}{22} .$

c) If exactly one of the first two is defective, 10 light bulbs remain, of which 2 are defective and 8 non-defective: $P (third non-defective) = \frac{8}{10} = \frac{4}{5} .$

d) The pack is accepted if the first two are non-defective, or if exactly one is defective and the third is non-defective: $P = \frac{6}{11} + \frac{9}{22} \cdot \frac{4}{5} = \frac{30}{55} + \frac{18}{55} = \frac{48}{55} .$

e) Conditional probability: accepted and one of the first two defective, divided by accepted: $P = \frac{\frac{9}{22} \cdot \frac{4}{5}}{\frac{48}{55}} = \frac{18/55}{48/55} = \frac{3}{8} .$

(ii) a) The sum of the probabilities is 1: $x = 1 \sum n k x = k \cdot \frac{n ( n + 1 )}{2} = 1 \Rightarrow k = \frac{2}{n ( n + 1 )} .$

b) The mean is $E (X) = x = 1 \sum n x \cdot k x = k x = 1 \sum n x^{2} = \frac{2}{n ( n + 1 )} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 n + 1}{3} .$

Answer: i) a) $\frac{6}{11}$ ; b) $\frac{9}{22}$ ; c) $\frac{4}{5}$ ; d) $\frac{48}{55}$ ; e) $\frac{3}{8}$ ; ii) a) $k = \frac{2}{n ( n + 1 )}$ ; b) $E (X) = \frac{2 n + 1}{3}$

Exercise 58 HardIB · P2 November 2001

Two women, Ann and Bridget, play a game which consists of taking turns to roll a fair six-sided die. The first woman to roll a six wins the game. Ann rolls first.

a) Calculate the probability that

(i) Bridget wins on her first roll.

(ii) Ann wins on her second roll.

(iii) Ann wins on her $n$ th roll.

b) Let $p$ be the probability that Ann wins the game. Show that $p = \frac{1}{6} + \frac{25}{36} p$ .

c) Calculate the probability that Bridget wins the game.

d) Suppose the game is played six times. Calculate the probability that Ann wins more games than Bridget.

Show answer

Answer: a) i) $\frac{5}{36}$ ; ii) $\frac{25}{216}$ ; iii) $(\frac{5}{6})^{2 n - 2} \cdot \frac{1}{6}$ ; c) $\frac{5}{11}$ ; d) $0.432$

Show working and answer

a) Each roll gives a six with probability $\frac{1}{6}$ and does not give one with $\frac{5}{6}$ .

(i) Ann fails and Bridget rolls a six: $\frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36} .$

(ii) Ann fails, Bridget fails, Ann rolls a six: $(\frac{5}{6})^{2} \cdot \frac{1}{6} = \frac{25}{216} .$

(iii) Before her $n$ th roll there are $n - 1$ failed rolls of Ann and $n - 1$ of Bridget: $(\frac{5}{6})^{2 n - 2} \cdot \frac{1}{6} .$

b) Ann wins if she rolls a six on her first roll, or if both fail and the game starts again in the same conditions: $p = \frac{1}{6} + \frac{5}{6} \cdot \frac{5}{6} \cdot p = \frac{1}{6} + \frac{25}{36} p .$

c) Solving: $p (1 - \frac{25}{36}) = \frac{1}{6} \Rightarrow \frac{11}{36} p = \frac{1}{6} \Rightarrow p = \frac{6}{11}$ . Since the game always ends, Bridget wins with probability $1 - \frac{6}{11} = \frac{5}{11} .$

d) In each game Ann wins with probability $\frac{6}{11}$ and Bridget with $\frac{5}{11}$ . Let $Y \sim B (6; \frac{6}{11})$ be the number of games Ann wins; she wins more games than Bridget if $Y \geq 4$ : $P (Y \geq 4) = k = 4 \sum 6 (k 6) (\frac{6}{11})^{k} (\frac{5}{11})^{6 - k} = \frac{765936}{1771561} \approx 0.432.$

Answer: a) i) $\frac{5}{36}$ ; ii) $\frac{25}{216}$ ; iii) $(\frac{5}{6})^{2 n - 2} \cdot \frac{1}{6}$ ; c) $\frac{5}{11}$ ; d) $0.432$

Exercise 59 HardIB · P2 TZ0 November 2009

In each round of two different games, Ying tosses three fair coins and Mario tosses two fair coins.

a) The first game consists of one round. If Ying gets more heads than Mario, she receives $5 from Mario. If Mario gets more heads than Ying, he receives $10 from Ying. If they get the same number of heads, then Mario receives $2 from Ying. Determine Ying's expected gain.

b) Now they play the second game, in which the winner is the player who gets the greater number of heads in one round. If they get the same number of heads, they play another round until there is a winner. Calculate the probability that Ying wins the game.

Show answer

Answer: a) expected gain of Ying $= 0$ $; b) $\frac{8}{11}$

Show working and answer

Number of heads: Ying (3 coins) $Y \sim B (3; \frac{1}{2})$ , Mario (2 coins) $M \sim B (2; \frac{1}{2})$ . There are $8 \cdot 4 = 32$ equally likely outcomes.

a) Counting the pairs $(Y, M)$ : $Y > M$ in 16 of them, $M > Y$ in 6 and a tie in 10. Hence $P (Y > M) = \frac{16}{32} = \frac{1}{2}, P (M > Y) = \frac{6}{32} = \frac{3}{16}, P (tie) = \frac{10}{32} = \frac{5}{16} .$ Expected gain of Ying: $5 \cdot \frac{1}{2} - 10 \cdot \frac{3}{16} - 2 \cdot \frac{5}{16} = \frac{40}{16} - \frac{30}{16} - \frac{10}{16} = 0.$

b) Tied rounds are repeated, so the probability that Ying wins is the probability that she wins a round given that there is a winner: $P = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{3}{16}} = \frac{8/16}{11/16} = \frac{8}{11} .$

Answer: a) expected gain of Ying $= 0$ $; b) $\frac{8}{11}$

Exercise 60 HardIB · P1 TZ0 November 2011

A target consists of three concentric circles of radii 1 m, 3 m and 5 m respectively, as shown in the figure.

Circular target with three concentric zones: a central white circle A of radius 1 m, a light grey ring B with outer radius 3 m and a dark grey ring C with outer radius 5 m

Nina throws an arrow at the target. The probability that the arrow sticks in the target is equal to $\frac{1}{2}$ . If the arrow sticks in the target, it does so at a random point on it. When a player throws, they are given ten points if the arrow sticks in region A, six points if it sticks in region B, and three points if it sticks in region C. Find the expected number of points that Nina obtains each time she throws an arrow at the target.

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Answer: expected number of points $= \frac{53}{25} = 2.12$

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Areas of the regions: A: $π \cdot 1^{2} = π$ ; B: $π (3^{2} - 1^{2}) = 8 π$ ; C: $π (5^{2} - 3^{2}) = 16 π$ ; total $25 π$ .

If the arrow sticks in the target, it lands at a random point, so $P (A) = \frac{1}{25} \cdot \frac{1}{2}, P (B) = \frac{8}{25} \cdot \frac{1}{2}, P (C) = \frac{16}{25} \cdot \frac{1}{2},$ and if it does not stick in the target (probability $\frac{1}{2}$ ) no points are scored. Hence: $E = 10 \cdot \frac{1}{50} + 6 \cdot \frac{8}{50} + 3 \cdot \frac{16}{50} + 0 \cdot \frac{1}{2} = \frac{10 + 48 + 48}{50} = \frac{106}{50} = \frac{53}{25} .$

Answer: expected number of points $= \frac{53}{25} = 2.12$

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