Sample spaces, tables and Venn diagrams
The sample space is the list of every possible outcome. For two coins it is {HH, HT, TH, TT}. For two dice, a 6 × 6 table gives 36 outcomes. If all outcomes are equally likely:
P(event) = number of outcomes you want ÷ total number of outcomes
A Venn diagram shows overlapping groups. The overlap is A and B (A ∩ B). Everything in either circle is A or B (A ∪ B). Outside both circles is neither.
Addition rule: P(A or B) = P(A) + P(B) − P(A and B). We subtract the overlap so it is not counted twice. If A and B cannot happen together (mutually exclusive), the overlap is 0.
Tree diagrams and the multiplication rule
A tree diagram draws each stage as a set of branches. Write the probability on every branch. The branches from one point always add up to 1.
- Multiply along a path to get the probability of that whole path: P(A then B) = P(A) × P(B after A).
- Add the end results of all paths you want. Example: P(one of each colour) = P(RB) + P(BR).
- All end results together add up to 1, which is a good check.
"At least one" questions are often quicker with 1 − P(none).
Independent and dependent events
Two events are independent if one does not change the chance of the other. Tossing a coin and rolling a die are independent. Picking a ball and putting it back before picking again is independent too. Then P(A and B) = P(A) × P(B).
Events are dependent if the first changes the second. Picking without replacement is dependent: after taking a red ball, there are fewer reds and fewer balls in total, so the second-stage fractions change.
Example: 3 red, 2 blue. With replacement P(RR) = 3/5 × 3/5 = 9/25. Without replacement P(RR) = 3/5 × 2/4 = 6/20 = 3/10.
Conditional probability
P(A | B) means "the probability of A given that B has happened". Knowing B shrinks the sample space to just B.
P(A | B) = P(A and B) ÷ P(B), or, with counts, (number in both) ÷ (number in B).
Example: of 30 students, 12 like cricket and 5 like both cricket and football. P(football | cricket) = 5 ÷ 12. Note this is not 5 ÷ 30: we only look at the cricket fans.
Rearranging gives the multiplication rule again: P(A and B) = P(B) × P(A | B). This is exactly what the second-stage branches of a tree diagram show.
Test for independence: A and B are independent if P(A | B) = P(A). Here P(F | C) = 5/12 ≈ 0.42 but P(F) = 18/30 = 0.6, so liking football and cricket are not independent.
Key formulas and definitions
- P(event) = favourable outcomes ÷ total outcomes (equally likely)
- P(A or B) = P(A) + P(B) − P(A and B)
- Independent: P(A and B) = P(A) × P(B)
- Any two events: P(A and B) = P(A) × P(B | A)
- P(A | B) = P(A and B) ÷ P(B)
- Independent ⇔ P(A | B) = P(A)
- P(at least one) = 1 − P(none)
Worked examples
1. Two fair dice are rolled. Find P(total = 7).
36 equally likely outcomes. Totals of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6. P = 6/36 = 1/6.
2. P(rain) = 0.3 on each of two days, independent. Find P(rain on both days).
0.3 × 0.3 = 0.09.
3. A bag has 4 green and 6 yellow sweets. Two are taken with replacement. Find P(both green).
4/10 × 4/10 = 16/100 = 4/25.
4. Same bag, without replacement. Find P(both green) and P(one of each).
P(GG) = 4/10 × 3/9 = 12/90 = 2/15. P(GY) = 4/10 × 6/9 = 24/90; P(YG) = 6/10 × 4/9 = 24/90. P(one of each) = 48/90 = 8/15.
5. In a class of 40, 22 study French, 15 study Spanish and 6 study both. Find P(French | Spanish).
Only look at the 15 Spanish students. 6 of them study French. P = 6/15 = 2/5.
6. P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.2. Are A and B independent? Find P(A | B).
P(A) × P(B) = 0.2 = P(A and B), so yes. P(A | B) = 0.2 ÷ 0.4 = 0.5 = P(A), which agrees.
7. A light is red with probability 0.6. If red, a driver is late with probability 0.3; if not red, 0.1. Find P(late).
Tree: P(red and late) = 0.6 × 0.3 = 0.18; P(not red and late) = 0.4 × 0.1 = 0.04. P(late) = 0.18 + 0.04 = 0.22.
Common mistakes
- Keeping the same second-stage fractions when there is no replacement. Take one away from the top and the bottom.
- Adding along a branch. Multiply along branches; add between finished paths.
- For P(A | B), dividing by the whole total instead of by the number in B.
- Thinking mutually exclusive means independent. Mutually exclusive events cannot both happen, so they affect each other.