IB Analysis and Approaches HL · Discrete random variables and the binomial distribution · Section 1 of 4

Discrete random variable and its probability distribution

3 worked examples · By Álvaro López

Key idea

A random variable turns the outcome of an experiment into a number. If it only takes isolated values (0, 1, 2, …) it is discrete, and it is described by a table giving the probability of each value. In the IB exam you are asked to complete tables with one unknown value, calculate cumulative probabilities and build the table from an experiment (draws, dice, spinners).

Discrete random variable

You toss three coins and want to know how many heads come up. The outcome of the experiment is a sequence such as ‘heads, tails, heads’, but you only record a number: 2. That number, which depends on chance, is a random variable. It is named with a capital letter, such as $X$ , and its specific values with a lower-case letter, such as $x$ .

The number of students who arrive late to a class, the points on a die or the red balls drawn from a bag are discrete. A person's height or the time a bus takes are continuous. In this topic only discrete variables are studied.

Probability distribution

The probability distribution of $X$ lists each possible value together with its probability $P (X = x)$ . It is written as a table:

$x$	$x_{1}$	$x_{2}$	$\dots$	$x_{n}$
$P (X = x)$	$P (X = x_{1})$	$P (X = x_{2})$	$\dots$	$P (X = x_{n})$

The reason is simple: the values of $X$ are mutually exclusive events (you cannot get 2 and 3 heads at the same time) and together they cover the whole sample space.

How to build the table for an experiment

When two dice or two spinners are involved, the most convenient method is a two-way table with all the outcomes equally likely: each cell is worth $\frac{1}{total number of cells}$ , and you only have to count how many cells give each value of $X$ .

Cumulative probabilities and unknown values

The probability that $X$ is at most $k$ is found by adding the table entries up to $k$ :

$P (X \leq k) = P (X = x_{1}) + P (X = x_{2}) + \dots + P (X = k)$

With strict inequalities you must check which values are included: $P (X < 3)$ does not include 3, whereas $P (X \geq 3)$ does. When it is shorter, use the complement: $P (X \geq 3) = 1 - P (X \leq 2)$ .

You can also combine this with conditional probability: $P (X \leq 2 ∣ X \geq 1) = \frac{P ( 1 \leq X \leq 2 )}{P ( X \geq 1 )}$ .

Worked examples

Example 1 Basic

Table with one unknown probability

The random variable $X$ is the number of calls received by a switchboard in one minute. Its probability distribution is

$x$	0	1	2	3
$P (X = x)$	$0.1$	$0.3$	$k$	$0.2$

a) Find the value of $k$ .

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

c) Find $P (X \leq 2 ∣ X \geq 1)$ .

Solution

a) The probabilities sum to 1: $0.1 + 0.3 + k + 0.2 = 1 \Rightarrow k = 0.4$

b) The values 2 and 3 are included: $P (X \geq 2) = 0.4 + 0.2 = 0.6$

c) By the formula for conditional probability, the numerator is $P (1 \leq X \leq 2) = 0.3 + 0.4 = 0.7$ and the denominator is $P (X \geq 1) = 1 - P (X = 0) = 0.9$ : $P (X \leq 2 ∣ X \geq 1) = \frac{0.7}{0.9} = \frac{7}{9}$

Answer: a) $k = 0.4$ ; b) $0.6$ ; c) $\frac{7}{9}$ .

Example 2 Medium

Three draws without replacement

A box contains 6 balls, 2 red and 4 blue. Three balls are drawn at random, without replacement. Let $X$ be the number of red balls obtained.

a) Write down the probability distribution of $X$ .

b) Find $P (X \geq 1)$ .

c) Find $P (X = 1 ∣ X \geq 1)$ .

Solution

a) There are $(3 6) = 20$ equally likely ways to choose 3 balls. The possible values are $0, 1, 2$ .

$X = 0$ : the 3 balls are blue, $(3 4) = 4$ ways, so $P (X = 0) = \frac{4}{20} = \frac{1}{5}$ .
$X = 1$ : one red and two blue, $(1 2) (2 4) = 2 \cdot 6 = 12$ ways, so $P (X = 1) = \frac{12}{20} = \frac{3}{5}$ .
$X = 2$ : both red and one blue, $(2 2) (1 4) = 4$ ways, so $P (X = 2) = \frac{4}{20} = \frac{1}{5}$ .

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

Check: $\frac{1}{5} + \frac{3}{5} + \frac{1}{5} = 1$ .

b) By the complement: $P (X \geq 1) = 1 - P (X = 0) = 1 - \frac{1}{5} = \frac{4}{5}$ .

c) $P (X = 1 ∣ X \geq 1) = \frac{P ( X = 1 )}{P ( X \geq 1 )} = \frac{3/5}{4/5} = \frac{3}{4}$

Answer: a) $P (X = 0) = \frac{1}{5}$ , $P (X = 1) = \frac{3}{5}$ , $P (X = 2) = \frac{1}{5}$ ; b) $\frac{4}{5}$ ; c) $\frac{3}{4}$ .

Example 3 Hard

Difference between two dice

Two fair six-sided dice are rolled. Let $X$ be the absolute difference between the two scores.

a) Construct the probability distribution of $X$ .

b) Find $P (X \geq 3)$ .

c) Find $P (X \geq 3 ∣ X \geq 1)$ .

d) What is the most likely value of $X$ ?

Solution

a) There are $36$ equally likely outcomes. $X$ takes the values $0, 1, 2, 3, 4, 5$ . Count the cells of the two-way table:

$X = 0$ : equal scores, 6 outcomes.
$X = 1$ : pairs such as $(1, 2)$ or $(2, 1)$ ; there are 5 pairs of consecutive values and each occurs in two orders, 10 outcomes.
$X = 2$ : 4 pairs, 8 outcomes.
$X = 3$ : 3 pairs, 6 outcomes.
$X = 4$ : 2 pairs, 4 outcomes.
$X = 5$ : only $(1, 6)$ and $(6, 1)$ , 2 outcomes.

$x$	0	1	2	3	4	5
$P (X = x)$	$\frac{6}{36}$	$\frac{10}{36}$	$\frac{8}{36}$	$\frac{6}{36}$	$\frac{4}{36}$	$\frac{2}{36}$

Check: $6 + 10 + 8 + 6 + 4 + 2 = 36$ .

b) $P (X \geq 3) = \frac{6 + 4 + 2}{36} = \frac{12}{36} = \frac{1}{3}$

c) With the condition $X \geq 1$ the denominator is $1 - P (X = 0) = \frac{30}{36}$ : $P (X \geq 3 ∣ X \geq 1) = \frac{12/36}{30/36} = \frac{12}{30} = \frac{2}{5}$

d) The largest value in the table is $\frac{10}{36}$ , which corresponds to $X = 1$ .

Answer: b) $\frac{1}{3}$ ; c) $\frac{2}{5}$ ; d) $X = 1$ .

In the exam: marks are given for showing the count of cases (or the two-way table) and checking that the sum is 1. In a conditional probability, write the numerator and the denominator separately before dividing.

Common mistakes

Summary

A random variable assigns a number to each outcome of an experiment; it is discrete if it takes isolated values.
Its probability distribution is the table of $P (X = x)$ for each value $x$ .
Always $\sum P (X = x) = 1$ and $0 \leq P (X = x) \leq 1$ : this helps you find an unknown value and check the table.
$P (X \leq k)$ adds the probabilities up to $k$ ; with inequalities, check whether the endpoint is included, and use the complement when it is shorter.
To build the table: define what $X$ counts, list its values, count the cases with a tree, a two-way table or combinations, and check that it sums to 1.
With a condition, $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ applies to values of $X$ just as to any event.

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