AS & A Level · Practice questions · Discrete probability distributions, E(X) and Var(X)

Table first, then expect.

Six original Cambridge-style questions on discrete probability distributions, expectation and variance, at Paper 5 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Values, probabilities, then sums.

Values: list every value $X$ can take.
Probabilities: trees, sample spaces or combinations; check they add to 1.
Expectation: $\text{E}(X) = \sum x p$.
Variance: $\text{E}(X^2) - [\text{E}(X)]^2$ with $\text{E}(X^2) = \sum x^2 p$.
Unknowns: one equation from $\sum p = 1$, another from the given $\text{E}(X)$.

Recall

[2 marks]

$\text{P}(X = x) = k(x + 1)$ for $x = 0, 1, 2, 3$. Find $k$ and $\text{P}(X \ge 2)$.

$k(1 + 2 + 3 + 4) = 1$, so $k = 0.1$. ✓
$\text{P}(X \ge 2) = 0.3 + 0.4 = 0.7$. ✓

Calculation

[3 marks]

$X$ takes values 1, 2, 3, 4 with probabilities 0.2, 0.3, 0.4, 0.1. Find $\text{E}(X)$ and $\text{Var}(X)$.

$\text{E}(X) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4$. ✓
$\text{E}(X^2) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6$. ✓
$\text{Var}(X) = 6.6 - 2.4^2 = 6.6 - 5.76 = 0.84$. ✓

Calculation

[5 marks]

Two fair dice are thrown. $X$ is the difference between the scores (larger minus smaller, or 0 if equal).

(a) Draw up the probability distribution of $X$. (b) Find $\text{E}(X)$ and $\text{Var}(X)$.

(a) Counting the 36 outcomes: $x = 0$: $\tfrac{6}{36}$; 1: $\tfrac{10}{36}$; 2: $\tfrac{8}{36}$; 3: $\tfrac{6}{36}$; 4: $\tfrac{4}{36}$; 5: $\tfrac{2}{36}$. ✓
(b) $\text{E}(X) = \dfrac{0 + 10 + 16 + 18 + 16 + 10}{36} = \dfrac{70}{36} = \dfrac{35}{18} = 1.94$ (3 s.f.). ✓
$\text{E}(X^2) = \dfrac{10 + 32 + 54 + 64 + 50}{36} = \dfrac{210}{36} = \dfrac{35}{6}$. ✓
$\text{Var}(X) = \dfrac{35}{6} - \left(\dfrac{35}{18}\right)^2 = 2.05$ (3 s.f.). ✓

Calculation

[4 marks]

$X$ takes values $-1, 0, 1, 2$ with probabilities $a$, 0.3, $b$, 0.2. Given that $\text{E}(X) = 0.5$, find $a$, $b$ and $\text{Var}(X)$.

$a + b = 0.5$ and $-a + b + 0.4 = 0.5$, so $b - a = 0.1$. ✓
$b = 0.3$, $a = 0.2$. ✓
$\text{E}(X^2) = 0.2 + 0 + 0.3 + 0.8 = 1.3$; $\text{Var}(X) = 1.3 - 0.25 = 1.05$. ✓

Analysis

[5 marks]

Three people are chosen at random from 4 women and 3 men. $X$ is the number of women chosen.

(a) Show that $\text{P}(X = 2) = \tfrac{18}{35}$. (b) Draw up the probability distribution of $X$. (c) Find $\text{E}(X)$ and $\text{Var}(X)$.

(a) $\dfrac{{}^{4}C_{2} \times {}^{3}C_{1}}{{}^{7}C_{3}} = \dfrac{6 \times 3}{35} = \dfrac{18}{35}$. ✓
(b) $\text{P}(X = 0) = \tfrac{1}{35}$, $\text{P}(X = 1) = \tfrac{4 \times 3}{35} = \tfrac{12}{35}$, $\text{P}(X = 2) = \tfrac{18}{35}$, $\text{P}(X = 3) = \tfrac{4}{35}$. ✓
(c) $\text{E}(X) = \dfrac{12 + 36 + 12}{35} = \dfrac{60}{35} = \dfrac{12}{7}$. ✓
$\text{E}(X^2) = \dfrac{12 + 72 + 36}{35} = \dfrac{24}{7}$; $\text{Var}(X) = \dfrac{24}{7} - \dfrac{144}{49} = \dfrac{24}{49}$. ✓

Analysis

[6 marks]

A bag contains 3 red and 2 blue discs. A player takes discs one at a time, without replacement, until a red disc is taken. $X$ is the number of discs taken.

(a) Find the probability distribution of $X$ and $\text{E}(X)$. (b) It costs $2.50 to play. The player receives $4 if $X = 1$, $1 if $X = 2$ and nothing if $X = 3$. Find the expected profit per game for the player, and comment.

(a) $\text{P}(X = 1) = \tfrac{3}{5}$; $\text{P}(X = 2) = \tfrac{2}{5} \times \tfrac{3}{4} = \tfrac{3}{10}$; $\text{P}(X = 3) = \tfrac{2}{5} \times \tfrac{1}{4} \times 1 = \tfrac{1}{10}$. ✓
$\text{E}(X) = 0.6 + 0.6 + 0.3 = 1.5$. ✓
(b) Expected receipts $= 0.6 \times 4 + 0.3 \times 1 + 0.1 \times 0 = 2.7$. ✓
Expected profit $= 2.7 - 2.5 = \$0.20$ per game. The player gains on average, so the game is not fair, and the organiser loses money in the long run. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.