Tree diagrams

Along and between

On a tree diagram, each pair of branches adds up to 1.

To find the probability of a path, multiply along it. If several paths give what you want, add their probabilities.

1. P(head) = 0.6 on each of two throws
2. P(two heads) = 0.6 × 0.6 = 0.36
3. P(exactly one head) = 0.6 × 0.4 + 0.4 × 0.6 = 0.48

Two events are independent if one happening does not change the probability of the other (like two separate coin throws). Then P(A and B) = P(A) × P(B). To test whether two events are independent, work out P(A) × P(B) and compare it with P(A and B): equal means independent. For "neither happens", multiply the two "not" probabilities: 0.7 × 0.7 = 0.49 when each event has probability 0.3.

With replacement (the item is put back), the probabilities are the same on every pick. For "both the same colour", add the paths: both red plus both blue.

When the second event depends on the first (such as being late depending on rain), the second set of branches has different probabilities after each first branch. Multiply along each route, then add the routes you want: P(rain) = 0.4, P(late | rain) = 0.5, P(late | no rain) = 0.1 gives P(late) = 0.4 × 0.5 + 0.6 × 0.1 = 0.26.

For three trials, count the orders. Exactly 2 heads in 3 throws can be HHT, HTH or THH: 3 orders, each with the same probability. With P(head) = 0.6: 3 × 0.6 × 0.6 × 0.4 = 0.432.