A-Level Maths 9709 / Topic 5.4 / AS & A Level

Every value, every probability.

A discrete random variable takes separate values, each with a probability. Drawing up its table from a situation, using the fact that the probabilities add to 1, and calculating $\text{E}(X)$ and $\text{Var}(X)$ are core Paper 5 skills.

Topic 5.4All discrete random variables lessons →

The key idea

The probabilities in a distribution add up to 1. $\text{E}(X) = \sum x\,p$ is the long-run mean, and $\text{Var}(X) = \sum x^2 p - \left[\text{E}(X)\right]^2$ measures the spread: the same 'mean of the squares minus the square of the mean' as for data.

Fig. 1: X is the larger score when two fair dice are thrown. P(X = x) = (2x − 1)/36, and the six probabilities add up to 36/36 = 1

Section 01

Probabilities add to one.

A probability distribution lists every value a discrete random variable $X$ can take, with its probability. Because one of the values must occur, $\sum \text{P}(X = x) = 1$. This single fact finds any unknown constant.

Worked example Finding a constant

The random variable $X$ has $\text{P}(X = x) = kx$ for $x = 1, 2, 3, 4$. Find $k$ and $\text{P}(X \ge 3)$.

$k(1) + k(2) + k(3) + k(4) = 1$, so $10k = 1$ and $k = 0.1$.
$\text{P}(X \ge 3) = \text{P}(X = 3) + \text{P}(X = 4) = 0.3 + 0.4$.

Answer: $k = 0.1$, $\text{P}(X \ge 3) = 0.7$

Your turn

$\text{P}(X = x) = \dfrac{x^2}{c}$ for $x = 1, 2, 3$. Find $c$.

Examiner trap

Forgetting that $\text{P}(X \ge 3)$ includes 3. 'At least 3' and 'more than 2' are the same event for whole-number values; 'more than 3' is not.

Section 02

Drawing up the table.

Most questions describe a situation and ask you to draw up the distribution. List the possible values first, then find each probability (with a tree, a sample space or combinations), and check that the total is 1.

Worked example Counters without replacement

A bag holds 4 red and 2 green counters. Two are taken at random without replacement. $X$ is the number of green counters taken. Draw up the probability distribution of $X$.

$X$ can be 0, 1 or 2.
$\text{P}(X = 0) = \tfrac{4}{6} \times \tfrac{3}{5} = \tfrac{12}{30} = \tfrac{2}{5}$.
$\text{P}(X = 1) = \tfrac{4}{6} \times \tfrac{2}{5} + \tfrac{2}{6} \times \tfrac{4}{5} = \tfrac{16}{30} = \tfrac{8}{15}$.
$\text{P}(X = 2) = \tfrac{2}{6} \times \tfrac{1}{5} = \tfrac{2}{30} = \tfrac{1}{15}$.
Check: $\tfrac{12 + 16 + 2}{30} = 1$.

Answer: $x$: 0, 1, 2 with probabilities $\tfrac{2}{5}, \tfrac{8}{15}, \tfrac{1}{15}$

Worked example The larger score on two dice

Two fair dice are thrown and $X$ is the larger of the two scores (or the common score if they are equal). Find the distribution of $X$.

$X = 1$ only for (1,1): $\tfrac{1}{36}$.
$X = 2$: (1,2), (2,1), (2,2): $\tfrac{3}{36}$.
In general $X = x$ when one die shows $x$ and the other shows $x$ or less: $2x - 1$ outcomes.
So the probabilities are $\tfrac{1}{36}, \tfrac{3}{36}, \tfrac{5}{36}, \tfrac{7}{36}, \tfrac{9}{36}, \tfrac{11}{36}$ (Fig. 1).

Answer: $\text{P}(X = x) = \dfrac{2x - 1}{36}$ for $x = 1, \dots, 6$

Examiner trap

Missing a route in $\text{P}(X = 1)$. One green can come first OR second: red-green and green-red are both needed.

Stage 1 · Learn

Check your understanding

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Quick check+10 XP

$X$ takes values 1, 2, 3, 4 with probabilities 0.2, 0.35, $p$, 0.15. Find $p$.

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$X$ takes values 1, 2, 3 with probabilities 0.2, 0.5, 0.3. Find $\text{E}(X)$.

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Which is the correct formula for $\text{Var}(X)$?

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Which of these could NOT be a probability distribution?

Section 03

Expectation and variance.

$\text{E}(X)$ is what the average value of $X$ would settle to over many trials. Multiply each value by its probability and add. For the variance, find $\text{E}(X^2) = \sum x^2 p$ the same way, then subtract the square of $\text{E}(X)$.

Worked example From a formula

For $\text{P}(X = x) = 0.1x$, $x = 1, 2, 3, 4$, find $\text{E}(X)$ and $\text{Var}(X)$.

$\text{E}(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3$.
$\text{E}(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 10$.
$\text{Var}(X) = 10 - 3^2$.

Answer: $\text{E}(X) = 3$, $\text{Var}(X) = 1$

Worked example From the counters table

Find $\text{E}(X)$ and $\text{Var}(X)$ for the number of green counters above.

$\text{E}(X) = 0 + 1 \times \tfrac{8}{15} + 2 \times \tfrac{1}{15} = \tfrac{10}{15} = \tfrac{2}{3}$.
$\text{E}(X^2) = 1 \times \tfrac{8}{15} + 4 \times \tfrac{1}{15} = \tfrac{12}{15} = \tfrac{4}{5}$.
$\text{Var}(X) = \tfrac{4}{5} - \tfrac{4}{9} = \tfrac{36 - 20}{45}$.

Answer: $\text{E}(X) = \tfrac{2}{3}$, $\text{Var}(X) = \tfrac{16}{45}$

Worked example Two unknowns

$X$ takes values 0, 1, 2, 3 with probabilities 0.1, $a$, $b$, 0.2. Given $\text{E}(X) = 1.7$, find $a$, $b$ and $\text{Var}(X)$.

Total 1: $a + b = 0.7$.
Expectation: $a + 2b + 0.6 = 1.7$, so $a + 2b = 1.1$.
Subtract: $b = 0.4$, then $a = 0.3$.
$\text{E}(X^2) = 0.3 + 4(0.4) + 9(0.2) = 3.7$; $\text{Var}(X) = 3.7 - 1.7^2 = 3.7 - 2.89$.

Answer: $a = 0.3$, $b = 0.4$, $\text{Var}(X) = 0.81$

Examiner trap

Stopping at $\text{E}(X^2)$, or writing $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)$. The expectation must be SQUARED before subtracting.

Section 04

Games and profit.

In a game, let $X$ be the profit (winnings minus the cost to play). $\text{E}(X) < 0$ means the player loses on average; $\text{E}(X) = 0$ is a fair game.

Worked example Is it worth playing?

It costs $1.50 to throw a fair die once. A six wins $5, a 4 or 5 wins $1, and anything else wins nothing. Find the expected profit and its variance.

Profit values: $5 - 1.5 = 3.5$ (prob. $\tfrac{1}{6}$), $1 - 1.5 = -0.5$ (prob. $\tfrac{2}{6}$), $-1.5$ (prob. $\tfrac{3}{6}$).
$\text{E}(X) = \dfrac{3.5 - 1 - 4.5}{6} = -\dfrac{2}{6} = -\dfrac{1}{3}$.
$\text{E}(X^2) = \dfrac{12.25 + 2(0.25) + 3(2.25)}{6} = \dfrac{19.5}{6} = 3.25$.
$\text{Var}(X) = 3.25 - \tfrac{1}{9} = 3.14$ (3 s.f.).

Answer: Expected profit $-\$0.33$ per game (a loss); variance 3.14

Examiner trap

Using the winnings instead of the profit. Expected winnings are $\tfrac{7}{6} = \$1.17$; you must still subtract the $1.50 stake.

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$X$ takes values 0, 1, 2, 3 with probabilities 0.1, $a$, $b$, 0.2, and $\text{E}(X) = 1.7$. Find $\text{Var}(X)$.

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Two counters are taken without replacement from a bag of 4 red and 2 green. $X$ is the number of green counters. Find $\text{Var}(X)$.

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$\text{P}(X = x) = kx^2$ for $x = 1, 2, 3$. Find $\text{E}(X)$.

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It costs $1.50 to throw a fair die. A six wins $5, a 4 or 5 wins $1, otherwise nothing. Find the expected profit per game.

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Six original Cambridge-style questions

Finding constants, drawing up tables from dice and selections, two unknowns from $\text{E}(X)$, variances and a game with profit.