A Level Maths 9709 · Paper 5 Probability & Statistics 1 · 5.3 · ~4 min read
Probability
Events, the addition rule, mutually exclusive and independent events, conditional probability, tree diagrams with and without replacement, and reversing a tree to find a "given" probability.
Sample spacesAddition ruleMutually exclusiveIndependent eventsConditional probabilityTree diagramsWithout replacementReversed conditioning
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Mutually exclusive events have P(A ∩ B) = 0.
- Independent events: P(A ∩ B) = P(A) × P(B). To show independence, test this equation with numbers.
- Conditional probability: P(A ǀ B) = P(A ∩ B)/P(B). "Given B" shrinks the sample space to B.
- Trees: multiply along branches, add the end results you want. Without replacement, the second-stage probabilities change.
By the end you can
- Use the addition rule and identify mutually exclusive events.
- Test for and use independence.
- Use tree diagrams, including problems without replacement.
- Calculate conditional probabilities, including the reverse of a tree.
1Combining events
The rules
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ; P(A ǀ B) = P(A ∩ B)/P(B) FORMULA LIST (MF19)
independent ⇔ P(A ∩ B) = P(A)P(B)- Mutually exclusive vs independent: different ideas. Mutually exclusive events can't both happen; independent events don't affect each other. (Two events with non-zero probabilities can't be both.)
- Venn diagrams help with two or three events; fill in the intersection first.
Worked example
P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7. Show that A and B are independent, and find P(A ǀ B).
- P(A ∩ B) = 0.4 + 0.5 − 0.7 = 0.2 M1
- P(A)P(B) = 0.2 = P(A ∩ B), so independent A1
- P(A ǀ B) = 0.2/0.5 = 0.4 (= P(A), as expected) B1
Common mistake"Independent because they don't overlap". That describes mutually exclusive events. Always justify independence with numbers.
2Tree diagrams
With and without replacement
Worked example
A bag holds 5 red and 3 blue counters. Two are taken without replacement. Find (a) P(same colour) and (b) P(first is red ǀ same colour).
- (a) (5/8)(4/7) + (3/8)(2/7) = 20/56 + 6/56 M1
- = 13/28 A1
- (b) P(RR)/P(same) = (20/56)/(26/56) M1
- = 10/13 A1
Worked example
Machine A makes 60% of a factory's items and 3% of its items are faulty; machine B makes the rest and 5% of its items are faulty. An item is chosen at random and is faulty. Find the probability it came from A.
- P(faulty) = 0.6 × 0.03 + 0.4 × 0.05 = 0.038 M1 A1
- P(A ǀ faulty) = 0.018/0.038 M1
- = 0.474 A1
Common mistakeUsing the branch value 0.03 as P(A ǀ faulty). That is P(faulty ǀ A). For the reverse, divide the joint probability by the total. This was the most common error on conditional probability in 2024.
Linking questionCounting methods are in Permutations and combinations; repeated independent trials lead to the binomial and geometric distributions in Discrete random variables.
Test yourself — cover the answers first
Quick check
- P(A) = 0.3. Find P(A′).
- A and B are mutually exclusive with P(A) = 0.2, P(B) = 0.5. Find P(A ∪ B).
- A and B are independent with P(A) = 0.2, P(B) = 0.5. Find P(A ∩ B).
- P(A ∩ B) = 0.12, P(B) = 0.4. Find P(A ǀ B).
- Two fair dice are thrown. Find P(total 7).
Exam-style
- P(A) = 0.6, P(B) = 0.3, P(A ∪ B) = 0.72. Are A and B independent?
- Two cards are drawn without replacement from 4 aces and 8 kings. Find P(at least one ace).
- A die is thrown twice. Given the total is 8, find P(at least one 6).
Show answers
- 0.7
- 0.7
- 0.1
- 0.3
- 6/36 = 1/6
- P(A ∩ B) = 0.18 = 0.6 × 0.3, so yes
- 1 − (8/12)(7/11) = 1 − 14/33 = 19/33
- totals of 8: (2,6), (3,5), (4,4), (5,3), (6,2); with a 6: 2, so 2/5
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