IB Applications and Interpretation SL · Discrete random variables and the binomial distribution · Section 1 of 3

Discrete random variable and probability distribution

3 worked examples · By Álvaro López

Key idea

A discrete random variable assigns a number to each outcome of a random experiment (heads obtained, points on a die, euros won). Its probability distribution is the table that says with what probability each value occurs. In the IB exam it almost always appears like this: you are given a table with an unknown value, asked to calculate it by imposing that the sum is 1, and then asked for probabilities. It is the basis of the expectation and of the binomial.

Random variable

A random experiment has outcomes that are not always numbers (heads or tails, a red or a black ball). To work with them mathematically, a number is assigned to each one, and that rule is the random variable.

The expression $P (X = x)$ is read "the probability that the variable $X$ takes the value $x$ ". Before calculating anything, write in words what $X$ counts and which values it can take: it is the first thing that gets marks.

Probability distribution

Definition · Probability distribution

The probability distribution of $X$ is the list of all its possible values with the probability of each one, usually in a table:

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

For a table to be a valid distribution, two conditions are needed.

This condition is the most useful tool in this section. If the table contains an unknown ( $k$ , $a$ …), write the sum equal to 1 and solve for it. Then check that all the resulting probabilities lie between 0 and 1: if one comes out negative, that solution is not valid.

Calculating probabilities from the table

As the values of $X$ are events that cannot occur together, the probability of a set of values is the sum of the probabilities of each one:

$P (X \geq x_{k}) = P (X = x_{k}) + P (X = x_{k + 1}) + \dots + P (X = x_{n})$

The inequality signs must be read carefully, because for a discrete variable it matters whether the endpoint is included: $P (X \geq 3)$ includes 3, $P (X > 3)$ does not, and for whole-number values $P (X > 3) = P (X \geq 4)$ .

The complement also helps: $P (X \geq 3) = 1 - P (X \leq 2)$ . It is shorter when the table has many values and only a few of them do not meet the required condition.

Bar chart of a discrete random variable with values 0 to 5 and probabilities 1/6, 5/18, 2/9, 1/6, 1/9 and 1/18; the bars for 3, 4 and 5 are highlighted with a brace showing that P(X ≥ 3) is the sum of those three bars

Worked examples

Example 1 Basic

A table with an unknown

A discrete random variable $X$ takes the values $1, 2, 3$ and $4$ , and its probability distribution is $P (X = x) = k x$ .

a) Calculate the value of $k$ .

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

c) Calculate $P (X < 3)$ .

Solution

a) The probabilities are $k, 2 k, 3 k, 4 k$ . As they add up to 1: $k + 2 k + 3 k + 4 k = 1 \Rightarrow 10 k = 1 \Rightarrow k = \frac{1}{10}$ The table becomes:

$x$	1	2	3	4
$P (X = x)$	$0.1$	$0.2$	$0.3$	$0.4$

b) $P (X \geq 3) = P (X = 3) + P (X = 4) = 0.3 + 0.4 = 0.7$ .

c) $P (X < 3) = P (X = 1) + P (X = 2) = 0.1 + 0.2 = 0.3$ . Check: $P (X < 3) = 1 - P (X \geq 3) = 1 - 0.7 = 0.3$ .

Example 2 Medium

The difference between two dice

Two rolls of a fair six-sided die (numbered 1 to 6) are made, and the variable $X$ is defined as the absolute value of the difference between the two results.

a) Construct the probability distribution of $X$ .

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

c) Calculate $P (1 \leq X \leq 3)$ .

Solution

a) The difference can be $0, 1, 2, 3, 4$ or $5$ . There are $36$ equally likely outcomes, which are counted in a two-way table:

$X = 0$ : the two dice are equal, $6$ cases.
$X = 1$ : pairs such as $(1, 2)$ or $(2, 1)$ . There are $5$ pairs of consecutive numbers and each can occur in two orders: $10$ cases.
$X = 2$ : $4$ pairs ( $1$ - $3$ , $2$ - $4$ , $3$ - $5$ , $4$ - $6$ ) in two orders: $8$ cases.
$X = 3$ : $3$ pairs in two orders: $6$ cases.
$X = 4$ : $2$ pairs in two orders: $4$ cases.
$X = 5$ : the pair $1$ - $6$ in two orders: $2$ cases.

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

Simplifying: $\frac{1}{6}, \frac{5}{18}, \frac{2}{9}, \frac{1}{6}, \frac{1}{9}, \frac{1}{18}$ . Check: $6 + 10 + 8 + 6 + 4 + 2 = 36$ , so they add up to 1.

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

c) $P (1 \leq X \leq 3) = \frac{10 + 8 + 6}{36} = \frac{24}{36} = \frac{2}{3}$ .

Example 3 Hard

Breakdowns of a machine

In a factory, the number of breakdowns $X$ suffered by a machine in one week takes the values $0, 1, 2$ and $3$ , with the following distribution, where $k > 0$ :

$x$	0	1	2	3
$P (X = x)$	$0.2$	$3 k$	$k^{2}$	$0.3$

a) Find the value of $k$ .

b) Calculate the probability that in one week there is at most one breakdown.

c) Given that there has been at least one breakdown in a week, calculate the probability that there has been exactly one.

d) The weeks are independent. Calculate the probability that there are no breakdowns in two consecutive weeks.

Solution

a) The sum of the probabilities must be 1: $0.2 + 3 k + k^{2} + 0.3 = 1 \Rightarrow k^{2} + 3 k - 0.5 = 0$ Using the quadratic formula (or the GDC equation solver): $k = \frac{- 3 \pm 9 + 2}{2} = \frac{- 3 \pm 11}{2}$ This gives $k = 0.158$ or $k = - 3.16$ . A probability cannot be negative ( $3 k < 0$ ), so the second is discarded and $k = \frac{- 3 + 11}{2} = 0.158$ Check: $3 k = 0.475$ and $k^{2} = 0.0251$ , both between 0 and 1, and $0.2 + 0.475 + 0.0251 + 0.3 = 1$ .

b) "At most one breakdown" is $X \leq 1$ : $P (X \leq 1) = P (X = 0) + P (X = 1) = 0.2 + 3 k = 0.2 + 0.475 = 0.675$

c) It is a conditional probability: $P (X = 1 ∣ X \geq 1)$ . The event $X \geq 1$ is the complement of $X = 0$ , so $P (X \geq 1) = 1 - 0.2 = 0.8$ . As $X = 1$ is contained in $X \geq 1$ : $P (X = 1 ∣ X \geq 1) = \frac{P ( X = 1 )}{P ( X \geq 1 )} = \frac{0.475}{0.8} = 0.594$

d) By independence, the probabilities of each week are multiplied: $P (no breakdowns in two weeks) = 0.2 \cdot 0.2 = 0.04$

In the exam: in part a) marks are given for writing the equation "sum $= 1$ " and for justifying why the negative root is discarded; if you give both roots without discarding either, you lose the last mark. Keep $k$ unrounded in the calculator for the following parts, and give the answers to 3 significant figures.

Common mistakes

Summary

A discrete random variable assigns a number to each outcome of chance and only takes isolated values that can be listed. It is written with a capital letter ( $X$ ) and its values with a lower-case letter ( $x$ ).
The probability distribution is the table with $P (X = x)$ for each value. It satisfies $0 \leq P (X = x) \leq 1$ and $\sum P (X = x) = 1$ .
If the table has an unknown, set up "sum $= 1$ ", solve, and discard the solutions that give probabilities outside $[0, 1]$ .
The probability of several values is the sum of their probabilities; the complement ( $1 - \dots$ ) often saves working.
Watch the endpoints: "at least" is $\geq$ , "more than" is $>$ , "at most" is $\leq$ , "fewer than" is $<$ .

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