Venn diagrams, tree diagrams, conditional probability, and the difference between mutually exclusive and independent.
Thirty students. Eighteen play football, twelve play tennis. How many play both is the only thing not fixed. Drag the overlap and watch every probability on the page follow it.
At one end they are mutually exclusive. Somewhere in the middle they are independent. Everywhere else, neither.
P(A ∪ B) = P(A) + P(B) − P(A &cup B). Add the two circles and you have counted the overlap twice, so take it off once.
P(A | B) = P(A ∩ B)P(B). Given B has happened, B is the new whole world, so divide by it.
Mutually exclusive and independent are not the same thing, and cannot both be true for events with non-zero probability. Mutually exclusive means the overlap is empty. Independent means knowing one tells you nothing about the other. If A and B are mutually exclusive then knowing A happened tells you B definitely did not, which is about as dependent as it gets.
| Mutually exclusive | Independent | |
|---|---|---|
| Test | P(A ∩ B) = 0 | P(A ∩ B) = P(A) × P(B) |
| In the diagram | the circles do not touch | a specific overlap, usually not a round number |
| P(A | B) | 0 | P(A) |
A bag holds 5 red and 3 blue counters. Two are drawn. Grow the tree and watch what happens to the second layer when the first counter does not go back.
Multiply along a branch, add between branches. "One of each" is two different paths, red then blue and blue then red, so you work out both and add them. Forgetting the second path halves the answer and is the commonest error on tree questions.
1. With 18 footballers, 12 tennis players and 7 who do both, out of 30 students, find P(F or T) to 3 decimal places.
2. Still with 7 playing both, find P(F | T) to 3 decimal places.
3. From the bag of 5 red and 3 blue, two drawn without replacement, find P(both red) to 4 decimal places.
Drawing the diagram. A Venn with all four regions filled in, or a tree with every branch labelled, is usually worth a mark on its own and prevents most of the errors that follow.
For conditional probability, dividing by the condition, not by the total. P(F | T) has 12 on the bottom, not 30.
On a tree without replacement, changing the second layer denominators. Leaving them at 8 is the single most expensive slip here.
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