Addition rule
How to read it
P( )
The probability of what is inside the brackets.
Or
A or B or both: at least one happens.
∪ means "or".
And
A and B: both happen.
∩ means "and".
Why subtract intersection?: If events overlap, we count intersection twice.
Subtract once to correct.
Example: football or tennis
In a class of 10 students, 5 play football, 4 play tennis and 2 play both. A student is chosen at random.
Find the probability that the student plays football or tennis.
Solution
- 5 + 4 − 2 = 7 students play football or tennis.
- Or use the rule with probabilities.
Final answer
7/10 = 0.7
Worked example
P(A)=0.4, P(B)=0.3, P(both)=0.1.
Find P(A or B).
Solution
- P(A or B)=0.4+0.3-0.1
- P(A or B)=0.6
Final answer
0.6 (subtract 0.1 to avoid double-counting).
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Multiplication rule and independence
How to read it
For independent events.
P( )
The probability of what is inside the brackets.
And
A and B: both happen.
∩ means "and".
Times
Multiply the two probabilities.
Independent events: Independent (one does not change the other): multiply as they are, e.g. 0.2 × 0.3.
Not independent: still multiply, but the second probability changes, e.g. 3 red and 2 blue, not put back: 3/5 × 2/4.
Worked example
A train is late with probability 0.2. On any day, independently, it rains with probability 0.3.
Find the probability that the train is late and it rains.
Solution
- P(late) = 0.2, P(rain) = 0.3. Independent: one does not change the other.
- P(late and rain) = 0.2 × 0.3 = 0.06
Final answer
0.06. Independent, so multiply.
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Conditional probability
How to read it
Given
B, given that A has happened.
| means "given".
Top
A and B both happen.
Bottom
A happens: the group you look at.
What it means: You already know A happened, so look only at the A group. Then ask: what fraction of that group is B?
E.g. 12 students are in the chess club and 3 of them wear glasses: P(glasses | chess club) = 3/12.
Worked example
A bag has 6 green and 4 yellow sweets. Two sweets are taken, one after the other, without putting the first back.
The first sweet is green. Find the probability that the second sweet is yellow.
Solution
- Given the first sweet is green: 9 sweets are left. All 4 yellow sweets are still there.
- Look only at the sweets that are left.
Final answer
4/9 ≈ 0.444
Show it: mutually exclusive or independent?
Mutually exclusive?: Means: they cannot happen together.
Test: is P(A ∩ B) = 0?
0 → mutually exclusive. More than 0 → not mutually exclusive.
Independent?: Means: one does not change the probability of the other.
Test: work out P(A) × P(B). Is it the same as P(A ∩ B)?
Same → independent. Different → not independent.
Mutually exclusive?
In a class of 28, 9 students play the piano, 12 play football and 4 do both. Are "plays the piano" and "plays football" mutually exclusive? Justify your answer.
Solution
- Mutually exclusive means they cannot happen together, so P(both) would be 0.
- 4 students do both.
Final answer
No: 4 students play both, so P(P ∩ F) = 4/28 ≠ 0. The events are not mutually exclusive.
Independent?
In a class of 20 students, 8 wear glasses, 10 play a sport and 4 do both. A student is chosen at random. Are "wears glasses" and "plays a sport" independent? Justify your answer.
Solution
- Find each probability from the question.
- Multiply them.
- Find P(both) from the question.
- Compare: they are the same.
Final answer
Yes: P(G) × P(S) = 0.2 = P(G ∩ S), so the events are independent.
Exam tip: "Justify" = the number that shows it AND the conclusion. "No" alone scores 0.