Probability

CIE A-Level Mathematics· Unit 4: Probability & Statistics 1, Topic 3: Probability· 25 min read

1. Core Definitions and Basic Probability★★☆☆☆⏱ 15 min

📘 Definition

Probability

$P (E)$

A measure of the likelihood of an event $E$ occurring, ranging from 0 (impossible) to 1 (certain). For equally likely outcomes, $P (E) = \frac{Number of favourable outcomes}{Total number of possible outcomes}$ .

Example:

Rolling a fair 6-sided die, the probability of rolling an even number is $\frac{3}{6} = 0.5$ .

When working with probability, the sample space describes all possible outcomes of an experiment, and an event is any subset of this sample space. A key basic concept is complementary probability: the complement of event $A$ (written $A^{'}$ ) is all outcomes not in $A$ , so $P (A^{'}) = 1 - P (A)$ .

📐 Worked Example

A bag contains 5 red, 3 blue and 2 green identical balls. A ball is drawn at random. What is the probability that the ball is not green?

1
Count the total number of balls in the bag:
2
$5 + 3 + 2 = 10$
3
Calculate the probability of drawing a green ball:
4
$P(\text{green}) = \frac{2}{10} = 0.2$
5
We want the probability of not drawing green, which is the complement of the event 'green':
6
$P(\text{not green}) = 1 - P(\text{green}) = 1 - 0.2 = 0.8$

2. Addition Rule and Mutually Exclusive Events★★☆☆☆⏱ 20 min

📘 Definition

Mutually Exclusive Events

Two events $A$ and $B$ are mutually exclusive (disjoint) if they cannot occur at the same time, meaning their intersection is empty: $A \cap B = \emptyset$ .

Example:

Drawing one card from a standard deck: getting a heart and getting a club are mutually exclusive.

The addition rule for two mutually exclusive events is simply $P (A \cup B) = P (A) + P (B)$ , because they share no outcomes. This, together with reasoning on a Venn diagram, is the approach CIE expects: split the sample space into non-overlapping regions ('only $A$ ', 'only $B$ ', 'both', 'neither') and add the parts you need. The same Venn-diagram partition also handles events that overlap, without double-counting the intersection. As optional enrichment, an overlap can instead be dealt with in one line by the general addition rule $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , but the 9709 syllabus does not require this formula — a Venn-diagram partition is always sufficient.

📐 Worked Example

A single card is drawn from a standard 52-card deck. Find the probability that it is a king or a queen.

1
Let $K$ be the event 'the card is a king' and $Q$ the event 'the card is a queen'. A single card cannot be both a king and a queen, so the events are mutually exclusive:
2
$P(K \cap Q) = 0$
3
There are 4 kings and 4 queens in the deck:
4
$P(K) = \frac{4}{52}, \quad P(Q) = \frac{4}{52}$
5
Because the events are mutually exclusive, the addition rule reduces to a direct sum:
6
$P(K \cup Q) = P(K) + P(Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

📐 Worked Example

In a group of 50 students, 25 play hockey, 20 play football, and 10 play both sports. Using a Venn diagram, find the probability that a randomly selected student plays neither sport.

1
Let $H$ be 'plays hockey' and $F$ be 'plays football'. Fill a Venn diagram from the centre outwards, starting with the overlap (students who play both):
2
$n(H \cap F) = 10$
3
Subtract the overlap from each total to get the 'only' regions:
4
$n(\text{only } H) = 25 - 10 = 15, \quad n(\text{only } F) = 20 - 10 = 10$
5
The three filled regions account for $15 + 10 + 10 = 35$ students, so the remaining students play neither sport:
6
$n(\text{neither}) = 50 - 35 = 15$
7
Convert the count to a probability:
8
$P(\text{neither}) = \frac{15}{50} = 0.3$

✓ Quick check

Test your understanding:

For two mutually exclusive events A and B, what is the value of $P (A \cap B)$ ?
- 0
- 1
- $P (A) P (B)$
- $P (A) + P (B)$
Reveal answer
0 —
Correct! Mutually exclusive events cannot overlap, so their intersection has probability 0.

3. Multiplication Rule and Independent Events★★★☆☆⏱ 20 min

📘 Definition

Independent Events

Two events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other. This means $P (A ∣ B) = P (A)$ and $P (B ∣ A) = P (B)$ , where $P (A ∣ B)$ is the conditional probability of $A$ given $B$ (defined fully in the next section).

Example:

Rolling two fair dice: the outcome of the first die does not affect the outcome of the second die.

The multiplication rule states that for any two events, $P (A \cap B) = P (A ∣ B) \times P (B)$ . If $A$ and $B$ are independent, $P (A ∣ B) = P (A)$ , so the rule simplifies to $P (A \cap B) = P (A) \times P (B)$ . In CIE exams, this formula is the standard test for independence: if $P (A \cap B) = P (A) P (B)$ , then $A$ and $B$ are independent.

📐 Worked Example

A bag contains 4 red and 6 blue balls. Two balls are drawn at random with replacement. Show that the events 'first ball is red' and 'second ball is red' are independent, and find the probability both are red.

1
Define $R_{1} =$ first ball is red, $R_{2} =$ second ball is red. With replacement, the total number of balls stays the same, so:
2
$P(R_1) = \frac{4}{10} = 0.4, \quad P(R_2) = \frac{4}{10} = 0.4$
3
Calculate the probability both are red using the multiplication rule:
4
$P(R_1 \cap R_2) = P(R_1) \times P(R_2) = 0.4 \times 0.4 = 0.16$
5
Check the independence condition: $P (R_{1}) \times P (R_{2}) = 0.16 = P (R_{1} \cap R_{2})$ , so the events are independent.

4. Conditional Probability and Tree Diagrams★★★☆☆⏱ 20 min

📘 Definition

Conditional Probability

$P (A ∣ B)$

The probability that event $A$ occurs given that event $B$ has already occurred, defined by $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ for $P (B) > 0$ . Rearranging recovers the multiplication rule $P (A \cap B) = P (A ∣ B) P (B)$ .

Example:

If a card drawn from a standard deck is known to be red, the probability it is a heart is $P (heart ∣ red) = \frac{13}{26} = 0.5$ , since 13 of the 26 red cards are hearts.

Conditional probability measures how the likelihood of one event changes once we know that another has occurred. For independent events the extra information makes no difference, so $P (A ∣ B) = P (A)$ ; for dependent events it does. Tree diagrams are the standard CIE tool: label each branch with a probability, make sure the branches leaving any point sum to 1, multiply along a path to reach a particular outcome, and add the paths that satisfy the condition. When sampling without replacement, the probabilities on the second set of branches are conditional on the first draw.

📐 Worked Example

A box contains 7 white and 3 black counters. Two counters are drawn at random without replacement. (a) Find the probability that both counters are black. (b) Given that the second counter is black, find the probability that the first counter was also black.

1
First draw: $P (black) = \frac{3}{10}$ and $P (white) = \frac{7}{10}$ — the first pair of branches on the tree.
2
The second draw depends on the first because there is no replacement. If the first counter was black, then 2 black and 7 white remain out of 9:
3
$P(\text{black}|\text{1st black}) = \frac{2}{9}, \quad P(\text{white}|\text{1st black}) = \frac{7}{9}$
4
(a) Multiply the probabilities along the 'black then black' path:
5
$P(\text{both black}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$
6
(b) Use $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ with $A$ = '1st black' and $B$ = '2nd black'. First find $P (2nd black)$ by adding the two paths that end in a black second counter ('white then black' and 'black then black'):
7
$P(\text{2nd black}) = \frac{7}{10} \times \frac{3}{9} + \frac{3}{10} \times \frac{2}{9} = \frac{21}{90} + \frac{6}{90} = \frac{27}{90} = \frac{3}{10}$
8
Then divide, using the 'both black' probability from part (a) as the numerator:
9
$P(\text{1st black}|\text{2nd black}) = \frac{P(\text{both black})}{P(\text{2nd black})} = \frac{1/15}{3/10} = \frac{6}{27} = \frac{2}{9}$

5. Common Pitfalls

Wrong move:

Confusing mutually exclusive and independent events

Why:

Students often assume these terms are interchangeable, but they describe completely different properties of events

Correct move:

Remember: mutually exclusive means $P (A \cap B) = 0$ (cannot occur together), independent means $P (A \cap B) = P (A) P (B)$ (one does not affect the other). Two events with non-zero probability cannot be both.

Wrong move:

Adding $P (A)$ and $P (B)$ directly when the two events can occur together

Why:

For events that overlap, adding the raw probabilities counts the shared outcomes twice and overstates $P (A \cup B)$ .

Correct move:

By default, reason on a Venn diagram: split the sample space into non-overlapping regions ('only $A$ ', 'only $B$ ', 'both', 'neither') and add only the parts you need. Substitute into the general formula $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ only when a value of $P (A \cap B)$ is actually given — the 9709 syllabus does not require this formula.

Wrong move:

Multiplying unadjusted probabilities when sampling without replacement

Why:

Students assume all combined events are independent and multiply raw probabilities, ignoring the changed sample space

Correct move:

For without replacement sampling, reduce the numerator and denominator of the second probability by 1 after drawing the first item.

Wrong move:

Double-counting the intersection region in Venn diagrams

Why:

Students add the full intersection value to both individual event regions instead of only counting it once in the union

Correct move:

When filling a Venn diagram, enter the intersection value first, then subtract the intersection from each individual event to get the 'only A' and 'only B' regions.

Wrong move:

Failing to check that a probability is between 0 and 1

Why:

Rushing through calculations leads to arithmetic errors that are easily caught

Correct move:

Always verify your final probability is between 0 and 1, and re-calculate if it is outside this range.

6. Quick Reference Cheatsheet

Concept	Formula
Complement rule	$P (A^{'}) = 1 - P (A)$
Addition rule (mutually exclusive)	$P (A \cup B) = P (A) + P (B)$
General addition rule (optional enrichment)	$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Conditional probability	$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$
General multiplication rule	$P (A \cap B) = P (A ∣ B) P (B)$
Multiplication rule (independent)	$P (A \cap B) = P (A) P (B)$
Test for independence	$P (A \cap B) = P (A) P (B)$
Mutually exclusive events	$P (A \cap B) = 0$

7. Frequently Asked

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot occur at the same time, so $P (A \cap B) = 0$ . Independent events do not affect each other's probability, so $P (A \cap B) = P (A) P (B)$ . Two events with non-zero probability cannot be both mutually exclusive and independent.

Going deeper

syllabusCIE 9709 A-Level Math SyllabusOfficial 9709 syllabus

What's Next

This foundational probability topic is required for all subsequent statistics topics in CIE A-Level 9709. The rules developed here — the addition and multiplication rules, mutually exclusive and independent events, and conditional probability with tree diagrams — are applied directly when you study the common probability distributions such as the binomial and normal distribution, which make up a large portion of the S1 exam. A solid understanding of the key rules and distinctions here will prevent common errors in almost all statistics questions you encounter.