Probability, Year 11: tree diagrams and independent events

8

提示Each new stage doubles the number of branch ends.

为什么After 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

提示Those branches cover every possible thing that can happen next.

为什么From 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

提示Moving along a path means one thing happens and then another.

为什么Multiply 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

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

为什么Heads 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

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

为什么0.42 + 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

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

为什么'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

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

为什么Each 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

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

为什么P(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

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

为什么For 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

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

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