Tree diagrams: probability with and without replacement

1. Draw one set of branches for the first event, one branch per outcome, each labelled with its probability.
2. From the end of each branch, draw branches for the second event.
3. Check that each set of branches adds up to 1.
4. At the end of each path, multiply to get the probability of that combination.
5. All the end results should add up to 1.

A bag has 3 red and 2 blue counters. A counter is taken, its colour noted, and it's put back. Then a second counter is taken.

Because the counter is replaced, the probabilities are the same both times: $P(R) = \tfrac{3}{5}$, $P(B) = \tfrac{2}{5}$.

The probability of one of each colour is $\tfrac{6}{25} + \tfrac{6}{25} = \tfrac{12}{25}$. Check: $9 + 6 + 6 + 4 = 25$, so the total is 1.

Same bag, but the first counter is not put back. Now there are only 4 counters for the second pick, and the probabilities depend on what was taken first.

$P(\text{both red}) = \tfrac{6}{20} = \tfrac{3}{10}$ and $P(\text{one of each}) = \tfrac{12}{20} = \tfrac{3}{5}$. Check: $6 + 6 + 6 + 2 = 20$.

Questions often signal this with "without replacement", or with situations like choosing two people from a group or eating two sweets from a bag.

"At least one" means one or more. The quickest way is usually:

In Example 1, $P(\text{at least one blue}) = 1 - P(\text{red, red}) = 1 - \tfrac{9}{25} = \tfrac{16}{25}$.

Tree diagrams also work when one event changes the chance of the next. The probability it rains is 0.3. If it rains, the probability Sam is late is 0.4; if it doesn't, it's 0.1.

Higher-tier questions may then ask "given that Sam was late, what's the probability it rained?" That's $\tfrac{0.12}{0.19} \approx 0.63$: the part of "late" that came from the rain branch.

• Forgetting to reduce both the top and the bottom of the fraction when there's no replacement.
• Adding along branches instead of multiplying.
• Only counting one order (red then blue) when the question allows either order.
• Simplifying fractions before you finish: keeping a common denominator makes adding easier.

1. The probability of winning a game is 0.6. Sam plays twice. Find P(wins both) and P(wins exactly one).
2. A box has 4 green and 6 yellow pens. Two are taken without replacement. Find P(both green).
3. Using the box in question 2, find P(both the same colour).

Answers: 1. $0.6 \times 0.6 = 0.36$; $0.6 \times 0.4 + 0.4 \times 0.6 = 0.48$. 2. $\tfrac{4}{10} \times \tfrac{3}{9} = \tfrac{12}{90} = \tfrac{2}{15}$. 3. $\tfrac{12}{90} + \tfrac{6}{10} \times \tfrac{5}{9} = \tfrac{12}{90} + \tfrac{30}{90} = \tfrac{42}{90} = \tfrac{7}{15}$.

More practice in our tree diagram questions.

• Fill in the tree diagram if one is given, even if the question doesn't ask: it earns method marks.
• Write the multiplication for each path you use, then the addition.
• Use the "add up to 1" check before moving on.
• Leave answers as fractions unless the question gives decimals.