Probability

Q: Which probability formulae do I get in the S1 exam?

The addition law and conditional probability rule are provided in the formula booklet. You must memorise the complementary probability rule and the independence multiplicative rule .

Edexcel International A-Level Mathematics· 2018 Specification (Issue 3) S1 §3.1-3.4· 75 min read

1. Core Probability Rules and Event Classification★★☆☆☆⏱ 15 min

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📘 Definition

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
$P(A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

2. Conditional Probability and Independent Events★★★☆☆⏱ 20 min

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📘 Definition

Conditional Probability

$P (B ∣ A)$

The probability that event $B$ occurs, given that event $A$ has already happened

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
$P(T) = P(M \cap T) + P(M' \cap T) = 0.21 + 0.14 = 0.35$

3. Venn Diagrams for Probability Problems★★★☆☆⏱ 15 min

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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
$\text{Probability} = \frac{5}{50} = 0.1$

4. Tree Diagrams, Sampling With and Without Replacement★★★★☆⏱ 20 min

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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 $P (r e d) = \frac{4}{10} = 0.4$ , second draw $P (r e d) = 0.4$ (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 $P (r e d) = \frac{3}{9} = \frac{1}{3}$
4
$P(both red) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133$

5. Common Pitfalls

Wrong move:

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:

This double counts outcomes that are in both events, often leading to a probability greater than 1, which is impossible

Correct move:

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 move:

Assuming events are independent without verification, e.g. using $P (A \cap B) = P (A) P (B)$ for sampling without replacement

Why:

Sampling without replacement makes events dependent, so the independence rule does not apply, leading to incorrect probabilities

Correct move:

Only use the independence multiplication rule if the question explicitly states events are independent, or you have proven they are independent

Wrong move:

Mixing up conditional probability order, e.g. calculating $P (A ∣ B)$ instead of $P (B ∣ A)$

Why:

Conditional probability is not symmetric: $P (B ∣ A) \neq = P (A ∣ B)$ unless $P (A) = P (B)$

Correct move:

Use the formula $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$ , ensuring the given event is the denominator

Wrong move:

Labelling Venn diagrams with total event counts instead of counts for outcomes only in that event (excluding intersections)

Why:

This leads to overcounting when summing values across circles, resulting in incorrect total counts that exceed the sample space size

Correct move:

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 move:

Forgetting to include the complement event when calculating total probability, e.g. omitting $P (M^{'} \cap T)$ when calculating $P (T)$

Why:

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 move:

Use the law of total probability: $P (B) = P (B ∣ A) P (A) + P (B ∣ A^{'}) P (A^{'})$ for complementary events $A$ and $A^{'}$

6. Quick Reference 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

7. Frequently Asked

Which probability formulae do I get in the S1 exam?

The addition law $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ and conditional probability rule $P (A \cap B) = P (A) P (B ∣ A)$ are provided in the formula booklet. You must memorise the complementary probability rule $P (A^{'}) = 1 - P (A)$ and the independence multiplicative rule $P (A \cap B) = P (A) P (B)$ .

Can I use tree or Venn diagrams instead of formulae for probability questions?

Yes, examiners accept fully labelled, clear diagram-based solutions for all S1 probability questions, as long as your working is easy to follow and all steps are shown.

What is the difference between sampling with and without replacement?

When sampling with replacement, each selection is independent so probabilities stay constant for all draws. When sampling without replacement, the total number of items decreases after each draw, so probabilities change between draws and events are dependent.

Going deeper

official_documentEdexcel IAL S1 SpecificationOfficial 2018 specification for S1 unit
official_documentEdexcel IAL Maths Formula BookletProvided in all S1 exams

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.