Every card in Probability, Year 11: tree diagrams and independent events
The whole deck, in order — so you can read it through before your child ever sees it.
← Back to Probability, Year 11: tree diagrams and independent events
- A tree diagram shows a coin being flipped three times, with two branches (heads and tails) at every stage. How many complete paths run from the start of the tree to its right-hand end? (number only)
8
HintEach new stage doubles the number of branch ends.
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.
- On a tree diagram, the probabilities on the branches that leave any one point add up to ____.
1
HintThose branches cover every possible thing that can happen next.
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.
- On a tree diagram, the first branch 'bus is late' has probability 0.2. From the end of that branch, the branch 'Sam misses registration' has probability 0.6. What is the probability of the outcome at the end of this path: the bus is late and Sam misses registration?
0.12
HintMoving along a path means one thing happens and then another.
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.
- A biased coin lands on heads with probability 0.6. It is flipped twice, and a tree diagram is drawn. What is the probability of getting exactly one head?
0.48
HintTwo different paths through the tree give one head. Find each, then combine.
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.
- A tree diagram for two stages has four paths. The probabilities at the ends of three of them are 0.42, 0.28 and 0.18. What is the probability at the end of the fourth path, and why?
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.
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.
- To find the probability that two independent events both happen, ____ their probabilities.
multiply
HintOnly a fraction of the times the first one happens will the second one happen too.
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.
- A fair six-sided dice is rolled twice. What is the probability of getting a six both times? Give a fraction in its simplest form.
1/36
HintThe second roll is not affected by the first, so combine the two chances for 'six and six'.
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).
- The probability that it rains on Saturday is 0.4 and the probability that it rains on Sunday is 0.5. Treat the two days as independent. What is the probability that it rains on at least one of the two days?
0.7
HintThe only way to miss 'at least one' is for both days to stay dry.
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.
- A and B are independent events. P(A) = 0.3 and P(A and B) = 0.12. What is P(B)? Give a decimal.
0.4
HintWrite the rule for 'A and B' as an equation and solve it.
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.
- A basketball player scores from each free throw with probability 0.8, and the throws are independent. She takes three free throws. What is the probability that she scores from all three? Give a decimal.
0.512
HintEach throw is unaffected by the others, so the rule for two events simply carries on to a third.
Why0.8 × 0.8 × 0.8 = 0.512. The multiplication rule for independent events extends to any number of events.
1★ GCSE-MATH-PRB-0001
2★ GCSE-MATH-PRB-0002
3★ GCSE-MATH-PRB-0003
4★ GCSE-MATH-PRB-0004
5★ GCSE-MATH-PRB-0005
6★ GCSE-MATH-PRB-0006
7★ GCSE-MATH-PRB-0007
8★ GCSE-MATH-PRB-0008
9★ GCSE-MATH-PRB-0009
10★ GCSE-MATH-PRB-0010
Keep what you learn
Here, nothing is saved. In your child’s own sky every card is scheduled — it comes back just before they’d forget it — and the professor who wrote it is one tap away.