Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — drawing without replacement
A bag contains 5 red and 3 blue balls. Two are drawn at random without replacement. Find the probability that both are red.
- 1First draw: P(red)=85Five reds out of eight balls.
- 2Second draw: one red has gone, so P(red)=74Both the top and the bottom drop by one. This is the step people get wrong.
- 3Multiply along the path: 85×74=5620Along a branch of a tree diagram you multiply.
- 4=145Simplify by 4.
Example 2 — the addition rule, and testing independence
For two events, P(A)=0.4, P(B)=0.5 and P(A∩B)=0.2. Find P(A∪B), and determine whether A and B are independent.
- 1P(A∪B)=P(A)+P(B)−P(A∩B)Subtract the overlap, or it is counted twice.
- 2=0.4+0.5−0.2=0.7Note this is under 1, as any probability must be.
- 3Test: P(A)×P(B)=0.4×0.5=0.2Independence is a comparison, not something you can see by looking.
- 4This equals P(A∩B)=0.2, so A and B are independent.Note they still overlap — independent does not mean mutually exclusive.
Example 3 — exactly one of two outcomes
From the same bag of 5 red and 3 blue balls, two are drawn without replacement. Find the probability that exactly one is red.
- 1"Exactly one red" happens two ways: red then blue, or blue then red.Missing the second path is the most common error in the topic.
- 2Red then blue: 85×73=5615After a red is removed, 3 blues remain out of 7.
- 3Blue then red: 83×75=5615After a blue is removed, 5 reds remain out of 7.
- 4Add the paths: 5615+5615=5630=2815Multiply along a path, add between paths.