Cambridge Syllabus Notes

Probability & Statistics 1 (S1) – Cambridge A‑Level Mathematics 9709

1. Basic Probability Rules

Probability tells us how likely an event is to happen. For any event A:

$$P(A)=\frac{\text{favourable outcomes}}{\text{total outcomes}}$$

Remember: 0 ≤ P(A) ≤ 1 – a probability can’t be negative or exceed 1.

Complementary rule: $P(A^c)=1-P(A)$ – the chance that A does NOT happen.
Addition rule (any two events): $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Mutually exclusive events: if A and B can’t happen together, $P(A\cup B)=P(A)+P(B)$.

Example (🎲 Dice roll): Rolling a fair die, what’s the probability of getting an even number? Outcomes: {2,4,6}. $$P(\text{even})=\frac{3}{6}=0.5$$

Outcome	Probability
1	$\frac{1}{6}$
2	$\frac{1}{6}$
3	$\frac{1}{6}$
4	$\frac{1}{6}$
5	$\frac{1}{6}$
6	$\frac{1}{6}$

2. Conditional Probability

Sometimes we know something has happened and want the probability of another event given that information.

$$P(A|B)=\frac{P(A\cap B)}{P(B)}\quad\text{if }P(B)>0$$

Card example (🃏): You draw two cards from a standard deck without replacement. What is the probability that the first card is an Ace given that the second card is an Ace?

There are 4 Aces in 52 cards. Probability second card is Ace: $P(B)=\frac{4}{52}=\frac{1}{13}$.
Probability both are Aces: $P(A\cap B)=\frac{4}{52}\times\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$.
Conditional probability: $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{221}}{\frac{1}{13}}=\frac{13}{221}\approx0.0588$.

3. Mutually Exclusive Events

Events that cannot both occur at the same time.

Rolling a die: getting a 2 or a 5. These two outcomes never happen together.
Drawing a red card or a black card from a deck – they’re mutually exclusive.

For mutually exclusive events, simply add the probabilities: $P(A\cup B)=P(A)+P(B)$.

4. Independent Events

Events where the outcome of one does not influence the other.

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

Coin toss example (🎯): Toss a fair coin twice.

$P(\text{heads on first toss})=\frac{1}{2}$.
$P(\text{heads on second toss})=\frac{1}{2}$.
Since the tosses are independent: $P(\text{heads on both})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

5. Exam Tips & Tricks

✔️ Read the question carefully. Look for words like “given”, “at least”, “exactly”, “mutually exclusive”, or “independent”.

✔️ Use a diagram or table. When in doubt, sketch a Venn diagram or a simple table to organise events.

✔️ Check your calculations. Remember that probabilities must be between 0 and 1; if you get something outside this range, re‑check.

✔️ Practice with real‑world examples. Think of everyday situations – dice, cards, coins – to build intuition.

Good luck! 🎓

Probability: rules, conditional probability, mutually exclusive and independent events