Tree diagrams in depth

Each pair of branches from one point adds up to 1. To find the probability of one route, multiply along it. When several routes give what you want, add the routes.

1. P(bus late) = 0.3 and P(train late) = 0.15, independent
2. P(bus on time) = 1 − 0.3 = 0.7 and P(train on time) = 1 − 0.15 = 0.85
3. late then on time: 0.3 × 0.85 = 0.255
4. on time then late: 0.7 × 0.15 = 0.105
5. P(exactly one late) = 0.255 + 0.105 = 0.36

A quick check: the four routes of a two-stage tree always add up to 1.

If the first item is not put back, the second pick has one fewer item in total, and one fewer of the colour already taken. A bag has 5 red and 3 blue counters, and two are taken:

1. red then blue: 5/8 × 3/7 = 15/56
2. blue then red: 3/8 × 5/7 = 15/56
3. P(one of each) = 15/56 + 15/56 = 30/56 = 15/28

With replacement every second branch would still be out of 8: 5/8 × 3/8 + 3/8 × 5/8 = 30/64.

"At least one" covers every route except one. Work out the route where it never happens and take it from 1: P(at least one) = 1 − P(none). This also works when the second branches depend on the first (Higher), as long as you follow the right branch.

1. P(pass maths) = 0.7; if she fails maths, P(pass science) = 0.4
2. P(neither) = 0.3 × 0.6 = 0.18
3. P(at least one pass) = 1 − 0.18 = 0.82

"Given that" cuts the tree down to the routes where the condition is true. Divide the probability you want by the probability of the condition.

1. 5 red and 3 blue, two taken without replacement
2. P(both red) = 5/8 × 4/7 = 20/56
3. P(both blue) = 3/8 × 2/7 = 6/56
4. P(same colour) = 20/56 + 6/56 = 26/56
5. P(both red given the same colour) = 20 ÷ 26 = 10/13

If the number of one colour is unknown, call it n, write each branch in terms of n and set the product equal to the probability you are given. Clear the fractions and you get a quadratic; reject the negative solution.