Probability 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
Addition and multiplication laws of probability	89	245	03–08, 12
Evaluating probabilities by enumeration or using permutations/combinations	69	210	01, 02, 06
Conditional probability	46	134	09, 10, 12
Mutually exclusive and independent events	36	90	03, 04, 11

Outcomes, events and equally likely outcomes

Syllabus requirement · §5.3

“

evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations

”

A few words are used in every question, so it is worth fixing them first.

An experiment is anything with an uncertain result: throwing a die, spinning a spinner, drawing a counter from a bag, choosing a student at random.
Each possible result is an outcome. Throwing one ordinary die has six outcomes: $1, 2, 3, 4, 5, 6$ .
The list of every possible outcome is the sample space.
An event is a collection of outcomes you are interested in, named by a capital letter. For one throw of a die, $A$ = "the score is even" is the collection $\{2, 4, 6\}$ .

The notation used throughout this note:

$P(A)$ is the probability that $A$ happens. It is always between $0$ (impossible) and $1$ (certain).
$A'$ , read " $A$ dash" or "not $A$ ", is the complement of $A$ : every outcome that is not in $A$ .
$A \cap B$ means " $A$ and $B$ ": the outcomes that are in both.
$A \cup B$ means " $A$ or $B$ (or both)".
$X$ (a capital letter for a number that depends on the outcome, such as "the number of heads") is a random variable, and $P(X = 2)$ means "the probability that $X$ comes out as $2$ ". Random variables are studied properly in the next note; here they are just a short way of naming an event.
In working, $\Longrightarrow$ means "which gives" and $\iff$ means "exactly when" (each statement implies the other).

Equally likely outcomes. When every outcome in the sample space has the same chance — a fair die, a fair spinner, an item chosen "at random" — a probability is nothing more than a count:

P(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes}}

For the even score on a die, $3$ of the $6$ outcomes are in $A$ , so $P(A) = \frac{3}{6} = \frac12$ .

The complement rule. Every outcome is either in $A$ or in $A'$ , never both. So if the sample space has $n$ outcomes and $A$ contains $k$ of them, $A'$ contains the other $n - k$ , and

P(A') = \frac{n - k}{n} = 1 - \frac{k}{n} = 1 - P(A)

This is used constantly: when an event has many outcomes, it is often quicker to count the few outcomes not in it and subtract from $1$ .

P(A') = 1 - P(A)

The complement rule. Holds for every event, equally likely outcomes or not.

Count outcomes that really are equally likely

The counting formula only works if the outcomes you count are equally likely. When two dice are thrown, the totals $2, 3, \dots, 12$ are not equally likely: a total of $7$ can happen in six ways, a total of $2$ in only one. The fix is always the same — count at the level where outcomes genuinely are equally likely. For two dice that is the $36$ ordered pairs (first die, second die): $(1,2)$ and $(2,1)$ are different outcomes. For three dice it is the $6 \times 6 \times 6 = 216$ ordered triples.

Why are ordered pairs equally likely? Imagine one die is red and the other blue. The red die is fair, so each of its $6$ faces is equally likely; whatever it shows, the blue die's $6$ faces are equally likely too. So each of the $6 \times 6$ (red, blue) pairs has the same chance, $\frac{1}{36}$ . The dice do not have to be different colours — it is just easier to see.

Probability by listing equally likely outcomes

1
Decide what one outcome is, keeping order: (first die, second die), (first spinner, second spinner, third spinner).
Ordered outcomes are the ones that are equally likely.
2
Count all the outcomes. For several independent dice or spinners, multiply the numbers of faces: two dice give $6 \times 6 = 36$ .
This is the denominator. A sample-space grid shows it at a glance for two dice or spinners.
3
List the outcomes in the event systematically (by first die, then second) and count them.
A system stops you missing or double-counting an outcome.
4
Divide, then simplify. If the event has many outcomes, count its complement instead and use $P(A) = 1 - P(A')$ .
The division is the definition of probability for equally likely outcomes.

A sample-space grid for two dice

Two fair four-sided dice, each numbered $1$ to $4$ , are thrown, and the two scores are added. Find the probability that the total is $5$ .

Show full working

Each of the 16 cells is one equally likely outcome (first die, second die), and shows its total. The four highlighted cells are the ones with total 5.

1
Decide what one outcome is. An outcome is an ordered pair (first die, second die), such as $(1, 4)$ .
Ordered pairs are equally likely; the totals are not, so we must not count totals.
2
Count all the outcomes. Each die has $4$ faces, so there are $4 \times 4 = 16$ equally likely outcomes — the $16$ cells of the grid.
For every one of the 4 results on the first die there are 4 results on the second.
3
List the outcomes with total $5$ , working through the first die in order: $(1,4),\ (2,3),\ (3,2),\ (4,1)$ That is $4$ outcomes.
Going through the first die 1, 2, 3, 4 in turn guarantees none is missed. (1, 4) and (4, 1) are different outcomes and both count.
4
Divide the favourable count by the total count. $P(\text{total} = 5) = \frac{4}{16}$
Favourable outcomes over total outcomes, because every outcome is equally likely.
5
Simplify. $\frac{4}{16} = \frac14$
Divide top and bottom by 4.

Answer

$\dfrac14$ .

Using the complement to save counting

The same two fair four-sided dice are thrown. Find the probability that at least one of the dice shows a $4$ .

Show full working

1
Name the event and its complement. Let $A$ = "at least one die shows $4$ ". Then $A'$ = "neither die shows $4$ ".
"At least one" covers three cases (first only, second only, both). Its complement is a single, simple case.
2
Count the outcomes in $A'$ . If neither die shows $4$ , each die is $1$ , $2$ or $3$ : $3 \times 3 = 9 \text{ outcomes}$
Three choices for the first die, three for the second — the top-left 3 × 3 block of the grid.
3
Find $P(A')$ . $P(A') = \frac{9}{16}$
9 of the 16 equally likely outcomes.
4
Use the complement rule. $P(A) = 1 - P(A') = 1 - \frac{9}{16} = \frac{7}{16}$
Check by counting directly: the fourth row and fourth column of the grid hold 4 + 4 − 1 = 7 cells (the cell (4, 4) is in both).

Answer

$\dfrac{7}{16}$ .

Whenever you see "at least one", try the complement "none" first.

Listing ordered outcomes for three dice

9709/53 M/J 2021 Q4(a)3 marks

Three fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time, repeatedly. For a single throw of the three dice, the score is the sum of the numbers on the top faces.

Find the probability that the score is 4 on a single throw of the three dice.

Show full working

1
Decide what one outcome is. An outcome is an ordered triple (first die, second die, third die).
Ordered triples are equally likely. A score is not.
2
Count all the outcomes. $6 \times 6 \times 6 = 216$
Six results for each of the three dice.
3
Find which numbers can make a score of $4$ . Every die shows at least $1$ , so the three numbers are at least $1 + 1 + 1 = 3$ . To reach $4$ , exactly one die must show $2$ and the other two must show $1$ .
Reasoning about the smallest possible total finds every case quickly, rather than trying triples at random.
4
List the ordered triples. The $2$ can be on the first, second or third die: $(2,1,1),\ (1,2,1),\ (1,1,2)$ That is $3$ outcomes.
Forgetting that the 2 can be on any of the three dice (and writing only 1 outcome) is the usual slip here.
5
Divide. $P(\text{score} = 4) = \frac{3}{216}$
Favourable over total.
6
Simplify. $\frac{3}{216} = \frac{1}{72}$
Divide top and bottom by 3.
7
The mark scheme's version. Each particular triple has probability $\left(\frac16\right)^3$ , and there are $3$ of them: $\left(\frac16\right)^3 \times 3 = \frac{3}{216} = \frac{1}{72}$
Same answer by the multiplication law of §04: the dice are independent, so one ordered triple has probability 1/6 × 1/6 × 1/6.

Answer

$\dfrac{1}{72}$ (or $0.0139$ ).

Common mistakes

Treating the totals $2, 3, \dots, 12$ of two dice as equally likely, and writing $P(\text{total} = 7) = \frac{1}{11}$
Count ordered pairs: $6$ of the $36$ give $7$ , so $P(\text{total} = 7) = \frac{6}{36} = \frac16$
Only the ordered pairs are equally likely.
Counting $(1, 2)$ and $(2, 1)$ as one outcome
Keep order: they are two different outcomes of the two dice
Dropping order undercounts the favourable outcomes but not the total, so the probability comes out too small.
Listing "at least one" cases one by one and missing the overlap
Use the complement: $P(\text{at least one}) = 1 - P(\text{none})$
The complement is a single case, so nothing can be missed or double-counted.

Your turn

Write down what one outcome is before you count anything.

1
Two fair six-sided dice are thrown and their scores are added. Find the probability that the total is $8$ .
Stuck? Show hint
There are 36 ordered pairs. Go through the first die from 1 to 6 and ask what the second die must be.
Show solution
1. 1
  Total outcomes: $6 \times 6 = 36$
  Ordered pairs (first die, second die).
2. 2
  Outcomes with total 8, first die in order: $(2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2)$ That is $5$ outcomes.
  A first die of 1 would need a 7 on the second die, which is impossible.
3. 3
  Divide: $P(\text{total} = 8) = \frac{5}{36}$
  5 favourable out of 36. The fraction does not simplify.
Answer
$\dfrac{5}{36}$ .
2
Three fair coins are thrown. Find the probability of obtaining exactly two heads.
Stuck? Show hint
List all $2 \times 2 \times 2 = 8$ ordered outcomes, such as HHT.
Show solution
1. 1
  All outcomes: $\text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}$ That is $2 \times 2 \times 2 = 8$ equally likely outcomes.
  Each coin has 2 results, and order matters.
2. 2
  Outcomes with exactly two heads: $\text{HHT, HTH, THH}$ That is $3$ outcomes.
  The single tail can be on any one of the three coins.
3. 3
  Divide: $P(\text{exactly two heads}) = \frac38$
  3 of 8 equally likely outcomes.
Answer
$\dfrac38$ .
3
Two fair six-sided dice are thrown and the two scores are multiplied. Find the probability that the product is an even number.
Stuck? Show hint
A product is odd only when both numbers are odd. Find P(odd product) first.
Show solution
1. 1
  Name the complement. The product is odd only if both scores are odd. So "even product" has complement "both scores odd".
  One even factor makes the whole product even, so the even case has many outcomes; the odd case has few.
2. 2
  Count the complement. Each die is $1$ , $3$ or $5$ : $3 \times 3 = 9 \text{ outcomes}$
  Three odd faces on each die.
3. 3
  Find $P(\text{odd product})$ . $\frac{9}{36} = \frac14$
  9 of the 36 ordered pairs.
4. 4
  Use the complement rule. $P(\text{even product}) = 1 - \frac14 = \frac34$
  Even and odd products cover every outcome between them.
Answer
$\dfrac34$ .
49709/53 O/N 2022 Q4(a)3 marks
Three fair 4-sided spinners each have sides labelled 1, 2, 3, 4. The spinners are spun at the same time and the number on the side on which each spinner lands is recorded. The random variable $X$ denotes the highest number recorded.

Show that $P(X = 2) = \frac{7}{64}$ .
Stuck? Show hint
The highest number is 2 when every spinner shows 1 or 2, but not all three show 1.
Show solution
1. 1
  Count all the outcomes. $4 \times 4 \times 4 = 64$ equally likely ordered triples.
  Four sides on each of three spinners.
2. 2
  Describe the event. The highest number is $2$ when every spinner shows $1$ or $2$ and at least one spinner shows $2$ .
  If any spinner showed 3 or 4 the highest number would be bigger; if all showed 1 it would be 1.
3. 3
  List the triples made of 1s and 2s with at least one 2: $222,\ 221,\ 212,\ 122,\ 211,\ 121,\ 112$ That is $7$ outcomes.
  One triple with three 2s, three with two 2s, three with one 2.
4. 4
  Divide. $P(X = 2) = \frac{7}{64}$ as required.
  A 'show that' needs the listed outcomes and the total 64 visible — the answer alone earns nothing.
5. 5
  A quicker check (the mark scheme's method 3), part 1. All three spinners show 1 or 2: $2 \times 2 \times 2 = 8$ of the $64$ outcomes, so $P(\text{highest} \leqslant 2) = \frac{8}{64}$
  Two allowed sides on each spinner. (The mark scheme writes this as (1/2)³, using the multiplication law of §04.)
6. 6
  Part 2. All three show 1: just $1$ outcome, $P(\text{highest} = 1) = \frac{1}{64}$
  Only (1, 1, 1).
7. 7
  Part 3. Subtract: $P(X = 2) = \frac{8}{64} - \frac{1}{64} = \frac{7}{64}$
  "Highest is at most 2" minus "highest is 1" leaves "highest is exactly 2".
Answer
$P(X = 2) = \dfrac{7}{64}$ .

Practise probability by counting outcomesReal past-paper questions · Evaluating probabilities by enumeration or using permutations/combinations

Checking your access…

Can you do all of these?

Count ordered outcomes (pairs, triples) — they are equally likely; totals are not
For a random selection, divide the favourable count by ⁿCᵣ, counting each group separately and multiplying; count top and bottom the same way
To decide whether events are exclusive, give a reason: an outcome in both (not exclusive), or why they cannot happen together
For "at least one", find P(none) and subtract from 1
Add only for cases that cannot happen together; multiply only along one sequence of stages
For "one of each", multiply one order by the number of orders (3! for three different results)
When trials have different probabilities, list every case (HHT, HTH, THH) and work each out separately
Without replacement, update both numerator and denominator after every draw; use the number of arrangements of the colour letters as the multiplier
Put a probability on every branch you draw, including 1, and label every outcome; an impossible branch (0) may be drawn or left out
For a transfer between bags, write down the second bag's contents on each branch before its probabilities
In a process that depends on the previous stage, choose each factor from the stage before; end branches where the process stops
For P(A | B), the event after "given" goes in the denominator; from a table, divide by that row or column total
Reading a tree backwards: denominator = all paths ending in the condition; numerator = those that also pass through the other event
To test independence, find P(A ∩ B) from the table, grid or tree, compare it with P(A) × P(B), and write a conclusion
With an unknown, build the probability in terms of it, equate to the given value, solve, and reject any impossible root with a reason

Now do the questions

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

Permutations and Combinations

Discrete Random Variables

Probability

Outcomes, events and equally likely outcomes

Counting selections with combinations

Mutually exclusive events and the addition law

Can you do all of these?

Outcomes, events and equally likely outcomes

The rest of this note

Can you do all of these?