# Probability

> Edexcel International A-Level Mathematics · IAL S1
> Source: https://www.owlsprep.com/study/edexcel-ial-math-s1-probability/

This guide covers all Edexcel IAL S1 (WST01) probability content for IAS exams, including core probability rules, conditional probability, independence, Venn/tree diagrams, and sampling with or without replacement.

**Prerequisites:** Basic fraction, decimal, and percentage arithmetic; Understanding of set notation (union, intersection, complement)

## Learning objectives

- Calculate basic probabilities using sample spaces and complementary event rules
- Apply addition and multiplication laws for mutually exclusive, non-exclusive, and independent events
- Solve conditional probability problems using given formulae and visual representations
- Construct and interpret Venn diagrams and tree diagrams for sampling with/without replacement

## Core Probability Rules and Event Classification

**Sample Space** — The complete set of all possible equally likely outcomes of a probability experiment

*Example:* Rolling a fair 6-sided die has sample space {1, 2, 3, 4, 5, 6}

The complement of event $A$, written $A'$, is the set of outcomes where $A$ does not occur. The rule $P(A') = 1 - P(A)$ must be memorised, as it is not provided in the formula booklet. Mutually exclusive events cannot occur at the same time, so $P(A \cap B) = 0$ for these events.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

The addition law above is given in your formula booklet. For mutually exclusive events, this simplifies to $P(A \cup B) = P(A) + P(B)$, as the intersection term is zero.

**Worked example:** A fair 6-sided die is rolled. Event $A$ = rolling an even number, Event $B$ = rolling a number greater than 4. Calculate (a) $P(A')$, (b) $P(A \cup B)$.

1. Identify sample space = {1,2,3,4,5,6}, all outcomes equally likely, so probability = number of favourable outcomes / total outcomes
2. $$P(A) = \frac{3}{6} = 0.5, \text{ so } P(A') = 1 - 0.5 = 0.5$$
3. Event $B$ = {5,6}, so $P(B) = \frac{2}{6} = \frac{1}{3}$. The intersection $A \cap B$ = {6}, so $P(A \cap B) = \frac{1}{6}$
4. $$P(A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$

> **Exam Tip**
>
> Always state any assumptions you make (e.g. 'die is fair so all outcomes are equally likely') to get full method marks, even if the question implies the assumption.

*Calculator:* allowed

## Conditional Probability and Independent Events

**Conditional Probability** — The probability that event $B$ occurs, given that event $A$ has already happened

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

*Example:* The probability it rains tomorrow given it rained today is an example of conditional probability

$$P(A \cap B) = P(A) \times P(B|A)$$

The formula above is provided in your formula booklet. Two events are independent if the occurrence of one does not change the probability of the other. For independent events, the following rules apply (you must memorise these):

- $P(B|A) = P(B)$
- $P(A|B) = P(A)$
- $P(A \cap B) = P(A) \times P(B)$

**Worked example:** The probability it rains on Monday is 0.3. If it rains on Monday, the probability it rains on Tuesday is 0.7. If it does not rain on Monday, the probability it rains on Tuesday is 0.2. Calculate $P(M \cap T)$ (rain both days) and $P(T)$ (rain on Tuesday).

1. Let $M$ = rain Monday, $T$ = rain Tuesday. $P(M) = 0.3$, so $P(M') = 1 - 0.3 = 0.7$
2. $$P(M \cap T) = P(M) \times P(T|M) = 0.3 \times 0.7 = 0.21$$
3. $$P(M' \cap T) = P(M') \times P(T|M') = 0.7 \times 0.2 = 0.14$$
4. $$P(T) = P(M \cap T) + P(M' \cap T) = 0.21 + 0.14 = 0.35$$

> **Exam Tip**
>
> If asked to prove two events are independent, calculate both sides of $P(A \cap B) = P(A)P(B)$ and show they are equal: do not just state they are independent.

*Calculator:* allowed

## Venn Diagrams for Probability Problems

Venn diagrams use circles inside a rectangle (representing the full sample space) to visually represent probabilities for up to 3 events. Always fill in the intersection value first, then the remaining parts of each circle, then the area outside all circles for the complement of the union of all events.

**Worked example:** In a group of 50 students, 22 study Physics, 25 study Chemistry, and 8 study neither subject. Calculate the probability a randomly selected student studies both Physics and Chemistry.

1. Let $P$ = study Physics, $C$ = study Chemistry. Total students = 50, number studying neither = 8, so $n(P \cup C) = 50 - 8 = 42$
2. $$n(P \cup C) = n(P) + n(C) - n(P \cap C)$$
3. $$42 = 22 + 25 - n(P \cap C) \implies n(P \cap C) = 47 - 42 = 5$$
4. $$\text{Probability} = \frac{5}{50} = 0.1$$

> **Exam Tip**
>
> Label all parts of your Venn diagram clearly, including the value for the area outside all circles, so examiners can follow your working easily.

*Calculator:* allowed

## Tree Diagrams, Sampling With and Without Replacement

Tree diagrams show sequences of events, with each branch labelled with its outcome and probability. Multiply probabilities along connected branches to get the probability of a full sequence of outcomes.

- **Sampling with replacement**: Items are returned to the pool after each draw, so all draws are independent and probabilities stay constant
- **Sampling without replacement**: Items are not returned after each draw, so total items decrease, probabilities change between draws, and events are dependent

**Worked example:** A bag contains 4 red and 6 blue balls. Two balls are drawn at random. Calculate the probability both are red if (a) sampling with replacement, (b) sampling without replacement.

1. (a) With replacement: First draw $P(red) = \frac{4}{10} = 0.4$, second draw $P(red) = 0.4$ (independent)
2. $$P(both red) = 0.4 \times 0.4 = 0.16$$
3. (b) Without replacement: After 1 red is drawn, 3 red and 6 blue balls remain, total 9, so second draw $P(red) = \frac{3}{9} = \frac{1}{3}$
4. $$P(both red) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133$$

> **Exam Tip**
>
> For sampling without replacement questions, always subtract 1 from both the total number of items and the number of favourable items for each subsequent draw, unless stated otherwise.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to subtract the intersection when applying the addition rule for non-mutually exclusive events, e.g. using $P(A \cup B) = P(A) + P(B)$ for overlapping events
  - Why it fails: This double counts outcomes that are in both events, often leading to a probability greater than 1, which is impossible
  - Correct: Always use the full addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ unless you have confirmed events are mutually exclusive
- **Wrong:** Assuming events are independent without verification, e.g. using $P(A \cap B) = P(A)P(B)$ for sampling without replacement
  - Why it fails: Sampling without replacement makes events dependent, so the independence rule does not apply, leading to incorrect probabilities
  - Correct: Only use the independence multiplication rule if the question explicitly states events are independent, or you have proven they are independent
- **Wrong:** Mixing up conditional probability order, e.g. calculating $P(A|B)$ instead of $P(B|A)$
  - Why it fails: Conditional probability is not symmetric: $P(B|A) \neq P(A|B)$ unless $P(A) = P(B)$
  - Correct: Use the formula $P(B|A) = \frac{P(A \cap B)}{P(A)}$, ensuring the given event is the denominator
- **Wrong:** Labelling Venn diagrams with total event counts instead of counts for outcomes *only* in that event (excluding intersections)
  - Why it fails: This leads to overcounting when summing values across circles, resulting in incorrect total counts that exceed the sample space size
  - Correct: Fill in the intersection value first, then subtract the intersection from the total event count to get the number of outcomes only in that event
- **Wrong:** Forgetting to include the complement event when calculating total probability, e.g. omitting $P(M' \cap T)$ when calculating $P(T)$
  - Why it fails: Most events can occur in multiple scenarios (e.g. rain on Tuesday when it rained Monday, or when it did not), so omitting any scenario gives an incomplete total
  - Correct: Use the law of total probability: $P(B) = P(B|A)P(A) + P(B|A')P(A')$ for complementary events $A$ and $A'$

## Cheatsheet

| Rule/Concept | Formula/Key Point | Formula Provided? |
| --- | --- | --- |
| Complementary Probability | $P(A') = 1 - P(A)$ | No (memorise) |
| Addition Law | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | Yes |
| Conditional Probability | $P(A \cap B) = P(A)P(B\|A)$ | Yes |
| Mutually Exclusive Events | $P(A \cap B) = 0$, $P(A \cup B) = P(A) + P(B)$ | No |
| Independent Events | $P(A \cap B) = P(A)P(B)$, $P(B\|A) = P(B)$ | No (memorise) |
| Sampling With Replacement | All draws independent, probabilities constant | No |
| Sampling Without Replacement | Draws dependent, probabilities change per draw | No |

## What's next

Now that you have mastered S1 probability, you are ready to move on to the next core S1 topics that build directly on these foundational skills. Probability is the basis for all statistical distributions you will encounter later in S1, including discrete random variables and the binomial distribution. You will also use these rules when interpreting correlation and regression results, to calculate the probability of observed patterns occurring by chance. Be sure to practice full exam-style questions that combine multiple probability concepts (e.g. Venn diagrams with conditional probability) to build confidence for your IAS exam.

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