Probability tree with branch probabilities

Why we multiply along a path

A fair coin is tossed, and then one ball is taken at random from a bag holding one red, one blue and one green ball. Heads happens in \(\dfrac12\) of all trials, and in \(\dfrac13\) of those trials the ball is green. So “heads and green” happens in \(\dfrac13\) of \(\dfrac12\) of all trials: \(\dfrac12\times\dfrac13=\dfrac16\). In the square below (area \(1\)) the coin splits the width in half, the ball splits each half into thirds, and the shaded rectangle is “heads and green”, with area \(\dfrac16\).

Probabilities on the branches

When outcomes are not equally likely, write each outcome’s probability on its branch. The branches leaving one junction cover every outcome of that stage, so their probabilities add up to \(1\). A spinner lands on red (\(R\)) with probability \(0.3\) and on blue (\(B\)) otherwise. It is spun twice; the spinner does not remember the first spin, so the second spin has the same probabilities.

Method

1. Draw one column of branches for each stage.
2. Write the probability on each branch; the branches from one junction add up to \(1\).
3. Multiply along a path to find the probability of that path.
4. Add the probabilities of the paths that make up the event: different paths cannot both happen.
5. For “at least one”, subtract the probability of “none” from \(1\).

Worked illustration

Both red: \(0.3\times0.3=0.09\). One of each colour: red then blue or blue then red, \(0.3\times0.7+0.7\times0.3=0.21+0.21=0.42\). At least one red: \(1-0.7\times0.7=1-0.49=0.51\).

Common errors

• Adding along a path: \(0.3+0.7\) is not the probability of red then blue.
• Taking only one order for “one of each colour”: \(0.21\) is red then blue only.

Check

• The path probabilities add up to \(1\): \(0.09+0.21+0.21+0.49=1\).