14F

Probability tree diagrams

Building probability trees with probabilities on the branches, multiplying along paths and adding paths, with and without replacement.

Building a probability tree

Example 1: without replacement●●

A bag holds $5$ red and $3$ blue marbles. Two are drawn without replacement. Find $P(\text{both red})$, $P(\text{both blue})$ and $P(\text{one of each})$.

$P(RR) = \dfrac58 \times \dfrac47 = \dfrac{5}{14}$

After a red is removed, 4 of the 7 left are red.

$P(BB) = \dfrac38 \times \dfrac27 = \dfrac{3}{28}$

$P(\text{one of each}) = \dfrac58 \times \dfrac37 + \dfrac38 \times \dfrac57 = \dfrac{15}{28}$

Two paths: RB and BR.

With replacement, the marble goes back, so the second stage is the same as the first: $P(RR) = \dfrac58 \times \dfrac58 = \dfrac{25}{64}$.

Example 2: a medical test●●●

$2\%$ of people have a condition. A test is positive for $95\%$ of people who have it and $10\%$ of people who don't. Find $P(\text{positive})$.

Path 1: has it and positive: $0.02 \times 0.95 = 0.019$.

Path 2: doesn't have it and positive: $0.98 \times 0.10 = 0.098$.

$P(\text{positive}) = 0.019 + 0.098 = 0.117$

Flashcards

Along a path

Multiply

Across paths

Add

Without replacement

The second-stage probabilities change.

Practice

From $5$ red and $3$ blue, two are drawn without replacement. Find $P(\text{both red})$.
Find $P(\text{both blue})$.
Find $P(\text{one of each colour})$.
Find $P(\text{both red})$ if the first marble is replaced.
The branches leaving one point of a probability tree:
To find the probability of an event made of several paths:
From $5$ red and $3$ blue without replacement, find $P(\text{at least one red})$.
In Example 2, find $P(\text{positive})$.
From $5$ red and $3$ blue without replacement, the first marble is red. Find the probability that the second is red.