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Probability Diagrams

Probability

Building one

  • A tree diagram sets out two or more stages, with a branch for each outcome at each stage
  • The syllabus states the convention these papers follow: outcomes at the ends of the branches, probabilities beside them
  • Each pair or group of branches from one point must add to 1, which fills in any missing probability
  • Questions almost always ask you to complete a partly-drawn tree first, and that alone carries marks
The layout of a two-stage tree diagram: the first experiment branches to A and not A with their probabilities beside the branches, each of those branches again to B and not B, and the four combined outcomes written at the ends as A and B, A and not B, not A and B, and not A and not B
Source: Tree diagrams by Save My Exams

Reading one

  • Follow one complete route across the diagram and multiply the probabilities you pass, which gives that path its own probability
  • Add between separate paths when more than one outcome satisfies the question
A tree diagram for two sets of traffic lights: the first branches green with probability 5 over 7 and red with 2 over 7, the second green with 8 over 9 and red with 1 over 9, and each of the four routes is multiplied along to give 40, 5, 16 and 2 out of 63
Source: Tree diagrams by Save My Exams
  • The probabilities of all the complete paths add to 1
  • For "without replacement", the second set of branches carries different numbers depending on which first branch you came along
Exam tip

Fill in every branch before calculating anything: each pair of branches from one point must add to 1, which supplies any probability the question has left out. Multiply along a route and add between separate routes, and leave the fractions uncancelled while you work so they combine easily at the end.

Worked example

Two counters without replacement

A bag holds 6 green counters and 4 white counters. Two counters are taken at random without replacement. Find the probability that both are green, that one of each colour is taken, and that at least one is white.

Solution:

  • There are 10 counters, so the first probabilities are 6/10 for green and 4/10 for white
  • If a green is taken first, 5 greens and 4 whites remain out of 9
  • If a white is taken first, 6 greens and 3 whites remain out of 9
  • P(both green) = 6/10 × 5/9 = 30/90, which simplifies to 1/3
  • One of each means green then white or white then green
  • P(one of each) = 6/10 × 4/9 + 4/10 × 6/9 = 24/90 + 24/90 = 48/90 = 8/15
  • At least one white is everything except both green, so subtract from 1
  • P(at least one white) = 1 − 30/90 = 60/90 = 2/3
  • Check: both green, both white and one of each come to 30/90 + 12/90 + 48/90 = 1