Probability, Year 11: tree diagrams and independent events

8

HintEach new stage doubles the number of branch ends.

The whyAfter one flip there are 2 ends, after two flips 2 × 2 = 4, and after three flips 2 × 2 × 2 = 8. Each path is one possible outcome, such as heads, tails, heads.

1

HintThose branches cover every possible thing that can happen next.

The whyFrom any point, exactly one of the branches must be followed, so their probabilities total 1. This lets you fill in a missing branch: if one of two branches is 0.3, the other is 0.7.

0.12

HintMoving along a path means one thing happens and then another.

The whyMultiply the probabilities along a path: 0.2 × 0.6 = 0.12. Sam misses registration on 60% of the 20% of days when the bus is late, which is 12% of all days.

0.48

HintTwo different paths through the tree give one head. Find each, then combine.

The whyHeads then tails: 0.6 × 0.4 = 0.24. Tails then heads: 0.4 × 0.6 = 0.24. The two paths are different outcomes, so add them: 0.24 + 0.24 = 0.48.

0.12 — the end-of-path probabilities cover every possible outcome, so they total 1

HintOne of the four routes through the diagram is certain to be followed.

The why0.42 + 0.28 + 0.18 = 0.88, and 1 − 0.88 = 0.12. Checking that all the final probabilities add to 1 is a quick test that a tree diagram has been filled in correctly.

multiply

HintOnly a fraction of the times the first one happens will the second one happen too.

The why'A and B' for independent events means P(A) × P(B). Adding is for 'A or B' when the two cannot happen together. Multiplying two probabilities gives a smaller number, which makes sense: both happening is less likely than either one.

1/36

HintThe second roll is not affected by the first, so combine the two chances for 'six and six'.

The whyEach roll gives a six with probability 1/6 and the rolls are independent, so P(six and six) = 1/6 × 1/6 = 1/36. On a 6 by 6 grid of outcomes, only one of the 36 squares is (6, 6).

0.7

HintThe only way to miss 'at least one' is for both days to stay dry.

The whyP(dry on both days) = 0.6 × 0.5 = 0.3, so P(rain on at least one) = 1 − 0.3 = 0.7. Adding the three paths with rain gives the same: 0.2 + 0.2 + 0.3 = 0.7.

0.4

HintWrite the rule for 'A and B' as an equation and solve it.

The whyFor independent events P(A and B) = P(A) × P(B), so 0.12 = 0.3 × P(B) and P(B) = 0.12 ÷ 0.3 = 0.4.

0.512

HintEach throw is unaffected by the others, so the rule for two events simply carries on to a third.

The why0.8 × 0.8 × 0.8 = 0.512. The multiplication rule for independent events extends to any number of events.