# Probability

> CIE A-Level Mathematics · 9709 Probability & Statistics 1
> Source: https://www.owlsprep.com/study/cie-9709-u4-probability/

This subtopic covers core probability concepts for CIE A-Level Mathematics, including fundamental rules for combined events, Venn/tree diagram problem-solving, and the key distinction between mutually exclusive and independent events, a foundation for all statistics topics.

**Prerequisites:** [Basic permutations and combinations](https://www.owlsprep.com/study/cie-9709-u4-permutations-and-combinations/); Basic set notation

## Learning objectives

- Calculate probabilities for single and combined events using core rules
- Distinguish between mutually exclusive and independent events
- Use Venn diagrams and tree diagrams to solve probability problems
- Apply the complement, addition and multiplication rules correctly

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 S1 past papers (2016–2025): **Probability** appears **91 times** in the last 10 years — **20.3%** of all S1 questions (91 of 438).

Most-tested forms: Conditional probability (60), Tree diagrams (30), Independent events (23), Multiplication rule (21), Sample space basics (15).

## Core Definitions and Basic Probability

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

*Notation:* $P(E)$

*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. $$P(\text{not green}) = 1 - P(\text{green}) = 1 - 0.2 = 0.8$$

> **tip**
>
> Always check your final probability is between 0 and 1. Any value outside this range indicates a calculation error.

## Addition Rule and Mutually Exclusive Events

**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. $$P(\text{neither}) = \frac{15}{50} = 0.3$$

**Check your understanding**

Test your understanding:

1. For two mutually exclusive events A and B, what is the value of $P(A \cap B)$?

   - 0
   - 1
   - $P(A)P(B)$
   - $P(A) + P(B)$

   *Why:* Correct! Mutually exclusive events cannot overlap, so their intersection has probability 0.

## Multiplication Rule and Independent Events

**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.

> **tip**
>
> Sampling with replacement always produces independent events, while sampling without replacement always produces dependent events, because the sample space changes after the first draw.

## Conditional Probability and Tree Diagrams

**Conditional Probability** — 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 $P(A \cap B) = P(A|B)\,P(B)$.

*Notation:* $P(A|B)$

*Example:* If a card drawn from a standard deck is known to be red, the probability it is a heart is $P(\text{heart}|\text{red}) = \frac{13}{26} = 0.5$, 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: $P(\text{black}) = \frac{3}{10}$ and $P(\text{white}) = \frac{7}{10}$ — 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 $P(\text{2nd black})$ 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. $$P(\text{1st black}|\text{2nd black}) = \frac{P(\text{both black})}{P(\text{2nd black})} = \frac{1/15}{3/10} = \frac{6}{27} = \frac{2}{9}$$

> **tip**
>
> On a tree diagram, multiply along the branches for a sequence of events and add the separate paths that meet the condition. Always check that the probabilities on the branches leaving the same point add up to 1.

## Common pitfalls

- **Wrong:** Confusing mutually exclusive and independent events
  - Why it fails: Students often assume these terms are interchangeable, but they describe completely different properties of events
  - Correct: 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:** Adding $P(A)$ and $P(B)$ directly when the two events can occur together
  - Why it fails: For events that overlap, adding the raw probabilities counts the shared outcomes twice and overstates $P(A \cup B)$.
  - Correct: 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:** Multiplying unadjusted probabilities when sampling without replacement
  - Why it fails: Students assume all combined events are independent and multiply raw probabilities, ignoring the changed sample space
  - Correct: For without replacement sampling, reduce the numerator and denominator of the second probability by 1 after drawing the first item.
- **Wrong:** Double-counting the intersection region in Venn diagrams
  - Why it fails: Students add the full intersection value to both individual event regions instead of only counting it once in the union
  - Correct: 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:** Failing to check that a probability is between 0 and 1
  - Why it fails: Rushing through calculations leads to arithmetic errors that are easily caught
  - Correct: Always verify your final probability is between 0 and 1, and re-calculate if it is outside this range.

## 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 | $P(A \cap B) = 0$ |

## 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.

- [Discrete random variables](https://www.owlsprep.com/study/cie-9709-u4-discrete-random-variables/)
- [Normal distribution](https://www.owlsprep.com/study/cie-9709-u4-normal-distribution/)

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