Discrete Random Variables Revision Notes — 9709 Paper 5

topic	marks/paper
Discrete Random Variables	13.2
The Normal Distribution	11.6
Probability	11.2
Permutations and Combinations	9.8
Representation of Data	9.3

sub-topic	parts	marks	sections
Binomial distribution B(n, p) and geometric distribution Geo(p)	110	278	07–09, 11–14
Probability distribution tables; E(X) and Var(X)	65	180	01–06
Expectation and variance of the binomial; expectation of the geometric	16	68	10, 11

Random variables and distribution tables from equally likely outcomes

Syllabus requirement · §5.4

“

draw up a probability distribution table relating to a given situation involving a discrete random variable $X$ , and calculate $\mathrm{E}(X)$ and $\mathrm{Var}(X)$ .

”

A random variable is a number whose value is decided by chance. "The number of heads when three coins are thrown", "the total score on two dice" and "the number of throws needed to get a 6" are all random variables: you cannot say in advance what the number will be, but you can say how likely each possible value is.

Two pieces of notation run through the whole topic:

a capital letter such as $X$ names the random variable itself — "the number of heads";
a lower-case letter such as $x$ stands for one particular value it can take — $x = 0$ , $x = 1$ , and so on. So $P(X = 2)$ is read "the probability that the number of heads is 2".

$X$ is discrete when its possible values can be listed one by one (usually whole numbers: $0, 1, 2, \ldots$ ), with gaps between them. Heights and times, which can take any value in a range, are continuous and belong to the Normal Distribution topic (syllabus 5.5).

A probability distribution table lists every value $X$ can take, with the probability of each:

$x$	$x_1$	$x_2$	$\cdots$	$x_n$
$P(X=x)$	$p_1$	$p_2$	$\cdots$	$p_n$

Every such table obeys two rules:

0 \leqslant P(X=x) \leqslant 1 \ \text{ for every } x, \qquad \sum P(X=x) = 1

The symbol $\sum$ (capital Greek sigma, also printed $\Sigma$ ) means "add up over every value of $x$ in the table". So the second rule says the probabilities in the bottom row always add up to exactly 1, because $X$ must take one of its listed values. It is the most useful check in the topic: after building any table, add the bottom row.

What "draw up the probability distribution table" earns marks for

Mark schemes give the first mark for a table with the correct list of $x$ -values, and the rest for the probabilities, each linked to the right $x$ -value. So always list every possible value first (including values such as $0$ that are easy to forget), then fill in the probabilities. Fractions do not need to be simplified; decimals must be exact or correct to at least 3 significant figures.

Drawing up a distribution table

1
List the possible values of $X$ . Read the rule that defines $X$ and write down every number it can produce.
This is the top row of the table, and it earns the first mark on its own.
2
Work out how each value can happen. For equally likely outcomes, list them all (a grid is best for two dice or spinners) and count; for unequal probabilities, see §02; for selections without replacement, see §03.
Each value of X is an event, and its probability comes from the outcomes that make up that event.
3
Write each probability in the table under its value of $x$ .
Keep fractions over the same denominator where you can — it makes the next step easy.
4
Check that the bottom row adds to 1.
If it doesn't, an outcome has been missed or counted twice. Fix it before using the table for anything else.

Equally likely outcomes: use a grid

When two fair dice or spinners are used together, every pair of results is equally likely. With a spinner of $a$ sides and one of $b$ sides there are $a \times b$ equally likely pairs, so

P(X = x) = \frac{\text{number of pairs that give } x}{a \times b}

The reliable way to count is a grid: one spinner along the top, the other down the side, and in each cell the value of $X$ that pair produces. Then count the cells holding each value.

Two spinners, X = the total score

A fair spinner $A$ has three sides numbered $1, 2, 3$ . A fair spinner $B$ has two sides numbered $1, 2$ . Both spinners are spun and $X$ is the sum of the two scores. Draw up the probability distribution table for $X$ .

Show full working

Spinner A down the side, spinner B along the top; each cell holds the total for that pair. Counting equal cells gives the table.

1
Count the equally likely pairs. Spinner $A$ has $3$ equally likely scores and spinner $B$ has $2$ , so there are $3 \times 2 = 6$ equally likely pairs.
The spinners are fair and independent, so every (A, B) pair has the same probability, 1/6.
2
Fill in the grid of totals. Row $A=1$ : $1+1=2$ , $1+2=3$ . Row $A=2$ : $2+1=3$ , $2+2=4$ . Row $A=3$ : $3+1=4$ , $3+2=5$ .
Writing the value of X in every cell is what stops a pair being missed.
3
List the possible values of $X$ . The totals in the grid are $2, 3, 4, 5$ .
The smallest total is 1 + 1 and the largest is 3 + 2, and every whole number between appears.
4
Count the cells for each value. $X=2$ : $1$ cell. $X=3$ : $2$ cells. $X=4$ : $2$ cells. $X=5$ : $1$ cell.
3 and 4 each happen in two different ways, so they are twice as likely as 2 or 5.
5
Divide each count by $6$ . $P(X=2)=\frac16, \quad P(X=3)=\frac26, \quad P(X=4)=\frac26, \quad P(X=5)=\frac16$
Leaving them over 6 keeps the check in the next step simple.
6
Check the sum. $\frac{1+2+2+1}{6} = \frac66 = 1 \ \checkmark$
The counts must add to the number of cells in the grid, 6.

Answer

$x$	$2$	$3$	$4$	$5$
$P(X=x)$	$\frac16$	$\frac26$	$\frac26$	$\frac16$

The same grid works for any rule — sum, product, difference, 'the higher score' — only the number written in each cell changes.

A rule with two cases, on two dice

9709/55 M/J 2025 Q1(a)3 marks

Two fair 6-sided dice with faces labelled $1, 2, 3, 4, 5, 6$ are thrown. The two scores are noted. The random variable $X$ is defined as follows.

If the two scores are equal, $X = 0$
If the scores are not equal, $X$ is the larger score minus the smaller score

Draw up the probability distribution table for $X$ .

Show full working

The 36 equally likely pairs, each labelled with X. The diagonal (equal scores) gives X = 0 in 6 cells; the bands on either side then give 10, 8, 6, 4 and 2 cells for X = 1 to 5.

1
Count the equally likely pairs. Each die has $6$ faces, so there are $6 \times 6 = 36$ equally likely ordered pairs.
(2, 5) and (5, 2) are different outcomes — the first die and the second die are different objects — so both appear in the grid.
2
Read the rule as one formula. For equal scores $X=0$ , and "larger minus smaller" is also $0$ when the scores are equal. So in every cell, $X = |\text{first} - \text{second}|$ , the difference between the two scores ignoring its sign.
Spotting that the two cases join up into one rule makes the grid quick to fill.
3
List the possible values of $X$ . The difference can be $0, 1, 2, 3, 4$ or $5$ .
The biggest possible difference is 6 − 1 = 5.
4
Count the cells for each value from the grid. $X=0$ : the diagonal, $6$ cells. $X=1$ : $5$ cells above the diagonal and $5$ below, $10$ cells. $X=2$ : $4+4=8$ . $X=3$ : $3+3=6$ . $X=4$ : $2+2=4$ . $X=5$ : $1+1=2$ .
Each band of equal differences runs parallel to the diagonal, one cell shorter on each side every time you move out.
5
Divide each count by $36$ . $P(0)=\frac{6}{36},\ P(1)=\frac{10}{36},\ P(2)=\frac{8}{36},\ P(3)=\frac{6}{36},\ P(4)=\frac{4}{36},\ P(5)=\frac{2}{36}$
The mark scheme accepts these unsimplified, or as 1/6, 5/18, 2/9, 1/6, 1/9, 1/18.
6
Check the sum. $\frac{6+10+8+6+4+2}{36} = \frac{36}{36} = 1 \ \checkmark$
The counts add to 36, the number of cells.

Answer

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

Part (b) of this question goes on to find E(X) and Var(X) from this table — it is worked in §05.

Dice and spinners with repeated numbers

Some questions use a die with faces such as $1, 2, 2, 3, 3, 3$ . It is still a fair die with six equally likely faces — the two faces marked $2$ are different faces that happen to show the same number. Give the grid one row (or column) per face, not per number, so the grid for two such dice is still $6 \times 6$ . Then $P(\text{score } 3) = \frac36$ , not $\frac13$ .

Shortcut: faces showing the same number can be grouped into one block of the grid, as long as the block keeps all its cells. For two dice with faces $1, 1, 1, 2, 2, 2$ and $1, 1, 2, 2, 3, 3$ , the pair "red 2, blue 3" is a block of $3 \times 2 = 6$ of the $36$ cells, so it has probability $\frac{6}{36}$ .

Common mistakes

Counting $(2, 5)$ and $(5, 2)$ as one outcome, so the grid for two dice has only $21$ cells
Use ordered pairs: two dice give $36$ equally likely outcomes
The 21 'unordered' outcomes are not equally likely — a double can happen one way, a non-double two ways.
Leaving a value out of the top row because it looks special, e.g. $X=0$
List every value the rule can produce before filling in probabilities
The first mark is for the correct set of x-values, so a missing value costs marks straight away.
Giving a die with faces $1, 1, 2, 2, 3, 3$ only three rows in the grid
One row per face: six rows, each with probability $\frac16$ — or one block of two rows per number
Three single rows would treat the die as having three faces with the wrong cell counts; each number must keep the 2 faces it really has.

Your turn

Draw the grid every time, even when you think you can see the counts.

19709/52 M/J 2021 Q4(a)3 marks
A fair spinner has sides numbered $1, 2, 2$ . Another fair spinner has sides numbered $-2, 0, 1$ . Each spinner is spun. The number on the side on which a spinner comes to rest is noted. The random variable $X$ is the sum of the numbers for the two spinners.

Draw up the probability distribution table for $X$ .
Stuck? Show hint
A 3 by 3 grid — give the first spinner one row per side, so the number 2 gets two rows.
Show solution
1. 1
  Count the pairs. $3 \times 3 = 9$ equally likely pairs.
  Each spinner has three equally likely sides.
2. 2
  Row for the side marked 1: $1+(-2)=-1$ , $\ 1+0=1$ , $\ 1+1=2$ .
  Add the first spinner's number to each of the second spinner's numbers.
3. 3
  Row for the first side marked 2: $2+(-2)=0$ , $\ 2+0=2$ , $\ 2+1=3$ .
4. 4
  Row for the second side marked 2: the same totals again, $0, 2, 3$ .
  Two different sides show 2, so this row appears twice — that is what makes totals involving 2 more likely.
5. 5
  Count each value. $X=-1$ : $1$ . $X=0$ : $2$ . $X=1$ : $1$ . $X=2$ : $3$ . $X=3$ : $2$ .
  The counts add to 9, the number of cells.
6. 6
  Divide by 9. $P(-1)=\frac19$ , $P(0)=\frac29$ , $P(1)=\frac19$ , $P(2)=\frac39$ , $P(3)=\frac29$ .
  Check: 1 + 2 + 1 + 3 + 2 = 9, so the probabilities add to 1.
Answer

$x$ $-1$ $0$ $1$ $2$ $3$
$P(X=x)$ $\frac19$ $\frac29$ $\frac19$ $\frac39$ $\frac29$
29709/53 O/N 2025 Q4(a)3 marks
A fair red spinner has 4 sides, numbered $1, 2, 3, 4$ . A fair blue spinner has 4 sides, numbered $0, 1, 2, 3$ . When a spinner is spun, the score is the number on the side on which it lands. The two spinners are spun at the same time.

The random variable $X$ denotes the higher of the two scores obtained. If the two scores are equal, then the value of $X$ is $0$ .

Draw up the probability distribution table for $X$ .
Stuck? Show hint
A 4 by 4 grid of 16 cells. Mark the cells where the two scores are equal first — those are all $X=0$ .
Show solution
1. 1
  Count the pairs. $4 \times 4 = 16$ equally likely pairs.
2. 2
  Equal scores give $X=0$ . The scores can be equal only at $1,1$ , $2,2$ and $3,3$ (red has no 0, blue has no 4): $3$ cells.
  Check the equal pairs before anything else, because the rule overrides 'the higher score' there.
3. 3
  Red score 1. Blue $0$ gives higher $=1$ . (Blue 1 is equal. Blue 2 and 3 give higher $2$ and $3$ .)
  Work across each row of the grid in turn.
4. 4
  Red score 2. Blue $0, 1$ give higher $=2$ ; blue $2$ is equal; blue $3$ gives $3$ .
5. 5
  Red score 3. Blue $0, 1, 2$ give higher $=3$ ; blue $3$ is equal.
6. 6
  Red score 4. Blue $0, 1, 2, 3$ all give higher $=4$ .
7. 7
  Count each value. $X=0$ : $3$ . $X=1$ : $1$ (red 1, blue 0). $X=2$ : red 1 blue 2, and red 2 blue 0 or 1: $3$ . $X=3$ : red 1 blue 3, red 2 blue 3, red 3 blue 0, 1, 2: $5$ . $X=4$ : $4$ .
  Check: 3 + 1 + 3 + 5 + 4 = 16.
Answer

$x$ $0$ $1$ $2$ $3$ $4$
$P(X=x)$ $\frac{3}{16}$ $\frac{1}{16}$ $\frac{3}{16}$ $\frac{5}{16}$ $\frac{4}{16}$
39709/53 O/N 2024 Q2(a)3 marks
A red fair six-sided dice has faces labelled $1, 1, 1, 2, 2, 2$ . A blue fair six-sided dice has faces labelled $1, 1, 2, 2, 3, 3$ . Both dice are thrown. The random variable $X$ is the product of the scores on the two dice.

Draw up the probability distribution table for $X$ .
Stuck? Show hint
Use the block shortcut from the callout: the 36-cell grid splits into blocks, one for each (red number, blue number) pair.
Show solution
1. 1
  Size of each block. Each red number is on $3$ faces and each blue number is on $2$ faces, so every (red number, blue number) block holds $3 \times 2 = 6$ of the $36$ cells.
  Grouping equal faces is fine because each block keeps all of its cells.
2. 2
  List the six blocks and their products. $1\times1=1$ , $\ 1\times2=2$ , $\ 1\times3=3$ , $\ 2\times1=2$ , $\ 2\times2=4$ , $\ 2\times3=6$ .
  Two rows of red numbers times three columns of blue numbers gives six blocks.
3. 3
  Count the cells for each product. $X=2$ appears in two blocks: $6 + 6 = 12$ cells. $X = 1, 3, 4, 6$ each appear in one block: $6$ cells.
  Products that come from different blocks are added together.
4. 4
  Divide by 36. $P(1)=\frac{6}{36}$ , $P(2)=\frac{12}{36}$ , $P(3)=\frac{6}{36}$ , $P(4)=\frac{6}{36}$ , $P(6)=\frac{6}{36}$ .
  Check: 6 + 12 + 6 + 6 + 6 = 36.
Answer

$x$ $1$ $2$ $3$ $4$ $6$
$P(X=x)$ $\frac{6}{36}$ $\frac{12}{36}$ $\frac{6}{36}$ $\frac{6}{36}$ $\frac{6}{36}$
49709/52 O/N 2024 Q3(a)3 marks
A fair coin and an ordinary fair six-sided dice are thrown at the same time. The random variable $X$ is defined as follows.
- If the coin shows a tail, $X$ is twice the score on the dice.
- If the coin shows a head, $X$ is the score on the dice if the score is even and $X$ is $0$ otherwise.
Draw up the probability distribution table for $X$ .
Stuck? Show hint
There are $2 \times 6 = 12$ equally likely (coin, dice) pairs. Make a 2-row grid: one row for tail, one for head.
Show solution
1. 1
  Count the pairs. $2 \times 6 = 12$ equally likely outcomes.
2. 2
  Tail row (twice the score): $2, 4, 6, 8, 10, 12$ .
  Each of these six outcomes has probability 1/12.
3. 3
  Head row (even score kept, odd score gives 0): dice $1,2,3,4,5,6$ give $0, 2, 0, 4, 0, 6$ .
4. 4
  Count each value across both rows. $X=0$ : $3$ . $X=2$ : $2$ (tail-1, head-2). $X=4$ : $2$ . $X=6$ : $2$ . $X=8, 10, 12$ : $1$ each.
  Check: 3 + 2 + 2 + 2 + 1 + 1 + 1 = 12.
Answer

$x$ $0$ $2$ $4$ $6$ $8$ $10$ $12$
$P(X=x)$ $\frac{3}{12}$ $\frac{2}{12}$ $\frac{2}{12}$ $\frac{2}{12}$ $\frac{1}{12}$ $\frac{1}{12}$ $\frac{1}{12}$

$x$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac19$	$\frac29$	$\frac19$	$\frac39$	$\frac29$

$x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{3}{16}$	$\frac{1}{16}$	$\frac{3}{16}$	$\frac{5}{16}$	$\frac{4}{16}$

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

$x$	$0$	$2$	$4$	$6$	$8$	$10$	$12$
$P(X=x)$	$\frac{3}{12}$	$\frac{2}{12}$	$\frac{2}{12}$	$\frac{2}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$

Practise probability distribution tablesReal past-paper questions · Probability distribution tables; expectation E(X) and variance Var(X)

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Can you do all of these?

List every possible value of X before any probability — the first mark is for the correct set of x-values
Use a grid of ordered pairs for two dice or spinners; give repeated faces their own rows
With biased coins, work out each scenario separately; only multiply by a number of orders when every order has the same probability
Without replacement: count selections with nCr for a handful, or multiply changing fractions for 'until' — never use the geometric formula
Add the bottom row of every table: it must be exactly 1
P(X = x) = kx²: substitute each x (a negative x squared is positive), sum to 1, solve for k, then give numbers
Var(X) = E(X²) − [E(X)]²; square x (not p) in E(X²); keep fractions exact
Two unknown probabilities: one equation from sum to 1, one from E(X) or the given condition
Binomial needs fixed n, two outcomes, constant p and independence; state conditions in context
Translate 'at least / more than / at most / fewer than' into a list of x-values, then take the shorter of the list and its complement
Write the unsimplified binomial terms before the decimals, keep 5–6 s.f. in each term, round only the answer
Smallest n: 1 − (1 − p)ⁿ > target; dividing by a negative log flips the inequality; round up
E(X) = np and Var(X) = np(1 − p); if both n and p are unknown, divide the variance by the mean
Geometric: no counting factor, r − 1 failures, E(X) = 1/p; for ranges ask 'how many trials must all fail?'
r-th success on trial n: choose r − 1 positions from the first n − 1 trials only, then multiply by p for the last trial
Two-stage questions: decide what one trial is, carry the earlier p unrounded

Now do the questions

183 real Paper 5 parts from 2021–2025, sorted by difficulty, with mark schemes

Probability

The Normal Distribution

Discrete Random Variables

Random variables and distribution tables from equally likely outcomes

Equally likely outcomes: use a grid

Distribution tables from independent events with different probabilities

Distribution tables from selections without replacement

Can you do all of these?

Random variables and distribution tables from equally likely outcomes

Equally likely outcomes: use a grid

The rest of this note

Can you do all of these?