IB Analysis and Approaches HL · Discrete random variables and the binomial distribution · Section 4 of 4

Binomial: cumulative probabilities, expected value and finding n or p

3 worked examples · By Álvaro López

Key idea

For the binomial distribution $X \sim B (n, p)$ the expected value and the variance are not calculated by summing: they are $n p$ and $n p (1 - p)$ . The typical exam questions ask for cumulative probabilities ( $P (X \leq k)$ , $P (X \geq k)$ ), “at least one” and finding $n$ or $p$ from given data. In Paper 2 they are solved with the graphic display calculator, but you must know how to turn the wording of the question into a correct inequality.

Expected value and variance of the binomial

The mean is intuitive: if each of the $n$ trials has probability $p$ of success, you expect $n p$ of them to be successes. For example, if 200 calls each have a 15% chance of being sales, you expect $200 \cdot 0.15 = 30$ sales. Note that $E (X)$ need not be an integer.

The variance is always smaller than the mean, because $n p (1 - p) < n p$ . When $p = \frac{1}{2}$ the spread is greatest, and when $p$ approaches 0 or 1, the bars become concentrated.

Cumulative probabilities

In practice a single exact value is rarely asked for, but a range: “at most 3”, “more than 5”, “between 2 and 4”. These are translated into $P (X \leq k)$ and calculated with the cumulative binomial distribution function of the GDC, which gives $P (X \leq k)$ .

The most common mistake is the “ $- 1$ ”: in $P (X \geq 4)$ the complement is $P (X \leq 3)$ , not $P (X \leq 4)$ . Write down which values are included before using the calculator.

“At least one” and finding n

This formula lets you solve a question that comes up repeatedly in exams: “how many trials are needed at least so that the probability of at least one success exceeds a given value?” One sets up $1 - (1 - p)^{n} > c$ and solves it:

In Paper 1: with logarithms, $(1 - p)^{n} < 1 - c$ , so $n > \frac{ln ( 1 - c )}{ln ( 1 - p )}$ (the inequality reverses because $ln (1 - p) < 0$ ).
In Paper 2: with a table of values of the function or by trying values on the GDC.

As $n$ is an integer, always round the value obtained up, unless it is already exact, and always check with the integer value.

Finding $p$ or $n$ from the mean and the variance

If the question gives $E (X)$ and $Var (X)$ , set up the system $n p = μ$ and $n p (1 - p) = σ^{2}$ . Dividing the second equation by the first gives $1 - p = σ^{2} / μ$ , from which $p$ follows and then $n = μ / p$ .

Worked examples

Example 1 Basic

Mean, standard deviation and the lower tail

A video game player obtains a rare item with probability $0.2$ in each game, independently. They play 10 games. Let $X$ be the number of rare items obtained.

a) Find $E (X)$ and $Var (X)$ .

b) Find the standard deviation $σ$ .

c) Find the probability that they obtain at most one item.

d) Find the probability that they obtain at least two items.

Solution

Here $X \sim B (10; 0.2)$ .

a) $E (X) = n p = 10 \cdot 0.2 = 2 Var (X) = n p (1 - p) = 10 \cdot 0.2 \cdot 0.8 = 1.6$

b) $σ = 1.6 \approx 1.26$

c) “At most one” is $X \leq 1$ : $P (X \leq 1) = P (X = 0) + P (X = 1) = 0. 8^{10} + 10 \cdot 0.2 \cdot 0. 8^{9} \approx 0.1074 + 0.2684 = 0.376$

d) “At least two” is the complement of the previous result: $P (X \geq 2) = 1 - P (X \leq 1) \approx 1 - 0.3758 = 0.624$

Answer: a) $E (X) = 2$ , $Var (X) = \frac{8}{5}$ ; b) $σ \approx 1.26$ ; c) $0.376$ ; d) $0.624$ .

Example 2 Medium

A 12-question test answered at random

A test has 12 questions with 4 options each, and only one option is correct. A student answers all the questions at random. Let $X$ be the number of correct answers.

a) Write down the distribution of $X$ and find the expected number of correct answers.

b) Find the probability of getting at least 4 answers correct.

c) Find the probability of getting at least one answer correct.

Solution

a) Each question is an independent trial with $p = \frac{1}{4}$ : $X \sim B (12, \frac{1}{4})$ . $E (X) = 12 \cdot \frac{1}{4} = 3$

b) “At least 4” is $X \geq 4$ , whose complement is $X \leq 3$ . With the cumulative binomial distribution, $P (X \leq 3) \approx 0.649$ , and $P (X \geq 4) = 1 - P (X \leq 3) \approx 1 - 0.649 = 0.351$

c) $P (X \geq 1) = 1 - P (X = 0) = 1 - (\frac{3}{4})^{12} \approx 1 - 0.0317 = 0.968$

Answer: a) $X \sim B (12, \frac{1}{4})$ , $E (X) = 3$ ; b) $0.351$ ; c) $0.968$ .

Example 3 Hard

Finding n and p, and a minimum number of trials

The random variable $X$ follows a binomial distribution with $E (X) = 12$ and $Var (X) = 7.2$ .

a) Find $n$ and $p$ .

b) Find $P (X = 12)$ and $P (X \leq 12)$ .

c) With this value of $p$ , $m$ independent trials are carried out. Find the smallest value of $m$ for which the probability of at least one success is greater than $0.99$ .

Solution

a) We set up $n p = 12$ and $n p (1 - p) = 7.2$ . Dividing the second equation by the first: $1 - p = \frac{7.2}{12} = 0.6 \Rightarrow p = 0.4$ and then $n = \frac{12}{0.4} = 30$ . Thus $X \sim B (30; 0.4)$ .

b) With the GDC (point and cumulative distributions): $P (X = 12) \approx 0.147 P (X \leq 12) \approx 0.578$ Note that the most likely value coincides with the mean, $n p = 12$ .

c) We need $P (at least one) = 1 - 0. 6^{m} > 0.99$ , that is, $0. 6^{m} < 0.01$ . Taking logarithms (and reversing the inequality when dividing by $ln 0.6 < 0$ ): $m > \frac{ln 0.01}{ln 0.6} \approx 9.015$ As $m$ is an integer, round up: $m = 10$ . Check: $0. 6^{9} \approx 0.01008 > 0.01$ is not enough, while $0. 6^{10} \approx 0.00605 < 0.01$ is.

Answer: a) $n = 30$ , $p = 0.4$ ; b) $0.147$ and $0.578$ ; c) $m = 10$ .

In the exam: marks are given for setting up the two equations in a), stating the probability asked for in b) and, in c), the inequality and the check of the integer. If you round $9.015$ down to $9$ , you lose the part.

Common mistakes

Summary

If $X \sim B (n, p)$ : $E (X) = n p$ , $Var (X) = n p (1 - p)$ , $σ = n p (1 - p)$ .
Translate the wording into $P (X \leq k)$ : “fewer than $k$ ” is $P (X \leq k - 1)$ ; “at least $k$ ” is $1 - P (X \leq k - 1)$ ; “more than $k$ ” is $1 - P (X \leq k)$ .
At least one success: $P (X \geq 1) = 1 - (1 - p)^{n}$ .
With $E (X)$ and $Var (X)$ : $1 - p = Var (X) / E (X)$ , so $p$ follows and then $n = E (X) / p$ .
To find the smallest $n$ in an inequality, solve with logarithms or with a GDC table, round up and check.
In Paper 2 use the point and cumulative binomial distribution functions of the GDC, and write the set-up with $n$ , $p$ and the inequality.

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Binomial: cumulative probabilities, expected value and finding n or p

Expected value and variance of the binomial

Cumulative probabilities

“At least one” and finding n

Finding p or n from the mean and the variance

Worked examples

Common mistakes

Summary

Want to work on it together?

Finding $p$ or $n$ from the mean and the variance