Tree diagrams and conditional probability — Expectation

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Tree diagrams and conditional probability — the key ideas

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Step 1 · Tree diagrams

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A tree diagram shows the outcomes of one event after another. Each branch is labelled with its probability, and the branches leaving any point must add up to 1. The second set of branches show probabilities given what happened first.

Step 2 · Multiply along, add between

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The probability of a particular route is found by multiplying along the branches ('and'). To find the probability of one of several routes, add the route probabilities ('or'). Check that all the end probabilities add up to 1.

Step 3 · Conditional probability

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$P(A \mid B)$ means 'the probability of $A$ given $B$ has happened': $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$. On a tree diagram, $P(A \cap B)$ is a route probability and $P(B)$ is the sum of all the routes in which $B$ happens.

Step 4 · Independence and replacement

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$A$ and $B$ are independent if $P(A \cap B) = P(A) \times P(B)$, equivalently $P(A \mid B) = P(A)$. Selecting with replacement gives independent events; without replacement the later probabilities change because the items left have changed.

Worked example

The probability that it rains is 0.3. If it rains, the probability the bus is late is 0.4; if it does not, 0.1. Find $P(\text{late})$ and $P(\text{rain} \mid \text{late})$.

Routes: rain and late = 0.3 × 0.4 = 0.12; no rain and late = 0.7 × 0.1 = 0.07.

$P(\text{late}) = 0.12 + 0.07 = 0.19$.

$P(\text{rain} \mid \text{late}) = \dfrac{0.12}{0.19} = \dfrac{12}{19} \approx 0.632$.

Test yourself

Past-paper style question

A bag contains 6 red counters and 4 blue counters. Two counters are taken at random without replacement.

(a) Complete a tree diagram for the two selections.
(b) Find the probability that both counters are the same colour.
(c) Given that the first counter is red, find the probability the second is blue. [5 marks]

Show the answer

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(a) First: R 6/10, B 4/10. After R: R 5/9, B 4/9. After B: R 6/9, B 3/9. (2 marks)

(b) $P(RR) + P(BB) = \frac{6}{10} \times \frac{5}{9} + \frac{4}{10} \times \frac{3}{9} = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}$. (2 marks)

(c) $P(\text{second blue} \mid \text{first red}) = \frac{4}{9}$. (1 mark)

Practice

Tree diagrams and conditional probability worksheet

Five short questions. Work through them, then reveal the mark scheme to check.

$A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.5$. Find $P(A \cap B)$.
A bag has 3 red and 2 blue counters. Two are taken with replacement. Find $P(\text{both red})$.
$P(A \cap B) = 0.12$ and $P(B) = 0.4$. Find $P(A \mid B)$.
$P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.25$. Are $A$ and $B$ independent?
A bag has 4 red and 6 blue counters. Two are taken without replacement. Find $P(\text{both red})$.

Mark scheme

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$0.3 \times 0.5 = $ 0.15.
$\frac{3}{5} \times \frac{3}{5} = $ 9/25.
$\dfrac{0.12}{0.4} = $ 0.3.
$P(A)P(B) = 0.5 \times 0.4 = 0.2$, which is not 0.25. So not independent.
$\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = $ 2/15.

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