Tree Diagrams: Edexcel GCSE Foundation maths revision notes

What a probability tree diagram shows

A probability tree diagram shows two events, one after the other, such as throwing a coin twice or playing two games. Each branch is labelled with an outcome and its probability.

Read a tree from left to right. The first pair of branches is the first event. Each of those branches splits into a pair for the second event, so a tree for two events has 4 routes from the start to the ends: heads then heads, heads then tails, tails then heads, and tails then tails.

In the exam you are usually asked to complete a tree diagram, for 2 marks, and then to work out the probability of one route through it, such as both things happening, for 2 more marks. Write the probabilities on a tree as fractions or decimals.

Completing a tree diagram

The branches that come out of the same point show everything that can happen at that stage, so their probabilities add up to 1:

P(\text{not } A)=1-P(A)

where A is an outcome, such as winning.

If the probability that a team wins is 0.35, the branch for "does not win" is 1-0.35=0.65. If the probability of red is \frac{3}{8}, the branch for "not red" is 1-\frac{3}{8}=\frac{5}{8}

The second pair of branches is drawn twice: once after each of the first branches. When the two events are independent, the first result does not change the chances for the second. Then both second pairs are the same, with the outcomes in the same order.

Check a finished tree: every branch has a probability, every pair adds up to 1, and each probability sits beside its own outcome.

Worked example 1: Completing a tree

Jess plays one game of tennis and then one game of squash. The probability that she wins at tennis is 0.55. The probability that she wins at squash is 0.4. The two results do not affect each other. Draw a probability tree diagram to show this information.

Tennis comes first, so it goes on the first pair of branches: win 0.55 and lose.
Your turn. What probability goes on the tennis branch for "lose"?

Your answer

1-0.55=0.45
Squash: win 0.4 and lose 1-0.4=0.6
The squash pair goes after both tennis branches, with the same probabilities each time, because the results do not affect each other.
Answer: tennis win 0.55, lose 0.45; squash after each tennis branch, win 0.4, lose 0.6

Both events: multiply along the branches

To find the probability of one route, such as "win then win", multiply the probabilities along its branches. For independent events A and B:

P(A\text{ and }B)=P(A)\times P(B)

Why multiply? Jess wins at tennis 0.55 of the time, and wins at squash in 0.4 of those times, so she wins both in 0.4 of 0.55 of the times: 0.55\times0.4=0.22

The four route probabilities always add up to 1, which is a useful check: 0.22+0.33+0.18+0.27=1

Without a calculator, multiply the digits, then count the decimal places. For 0.2\times0.3, 2\times3=6 and there are two decimal places, so the answer is 0.06. For fractions, multiply the tops and multiply the bottoms: \frac{2}{3}\times\frac{1}{4}=\frac{2}{12}=\frac{1}{6}

Worked example 2: Both and neither

Mo plants a sunflower seed and a tomato seed. The probability that the sunflower seed grows is 0.8. The probability that the tomato seed grows is 0.3. The two seeds do not affect each other. Work out the probability that both seeds grow. Work out the probability that neither seed grows.

Fails to grow: sunflower 1-0.8=0.2, tomato 1-0.3=0.7
Both grow: follow "grows" then "grows". 0.8\times0.3=0.24, because 8\times3=24 with two decimal places.
Your turn. Neither grows means "fails" then "fails". Work out 0.2\times0.7

Your answer

0.2\times0.7=0.14
Answer: both grow 0.24; neither grows 0.14

Finding the route the question asks for

The question tells you which route to follow. Read its words carefully, find the one route that fits, then multiply along it.

For two games that are each won or lost: - Both games won: win then win. - Neither game won, or "does not win either game": lose then lose. - Wins game 1 and loses game 2: win then lose. - Loses game 1 and wins game 2: lose then win.

Win then lose and lose then win are different routes. Find the first event's branch on the left of the tree, then follow the second event's branch that comes after it.

Worked example 3: One happens and the other does not

Ola goes fishing on Saturday and on Sunday. The probability that she catches a fish on Saturday is 0.7. The probability that she catches a fish on Sunday is 0.15. The two days do not affect each other. Work out the probability that Ola catches a fish on Saturday and does not catch a fish on Sunday.

No fish: Saturday 1-0.7=0.3, Sunday 1-0.15=0.85
The route is "fish" on Saturday, then "no fish" on Sunday: 0.7 then 0.85
Your turn. Work out 0.7\times0.85

Your answer

0.7\times0.85=0.595, because 7\times85=595 with three decimal places.
Answer: 0.595

Probabilities from bags, spinners and dice

When the probabilities come from things you can count, write each one as a fraction: the number of that kind over the total number. A bag of 4 red and 3 blue counters gives red \frac{4}{7} and blue \frac{3}{7}

A fair spinner or a fair dice makes every number equally likely. On a fair 8-sided spinner numbered 1 to 8, the probability of landing on 8 is \frac{1}{8} and the probability of not landing on 8 is \frac{7}{8}

Keep these as fractions. As a decimal, \frac{1}{3} is 0.333\ldots and never ends, so rounding it makes your answer inaccurate.

If a counter is taken from a bag and put back before the second pick, the bag is the same both times, so the second branches are the same as the first.

You can leave fractions such as \frac{3}{6} as they are, on the tree and in your answer. You do not need to simplify them unless the question asks you to.

Worked example 4: A bead from each of two bags

Bag A contains 3 red beads and 2 white beads. Bag B contains 1 red bead and 3 white beads. Kit takes at random a bead from bag A and a bead from bag B. Work out the probability that both beads are white.

Bag A has 3+2=5 beads: red \frac{3}{5}, white \frac{2}{5}
Your turn. Bag B has 4 beads. What is the probability that the bead from bag B is white?

Your answer

3 of the 4 beads in bag B are white: white \frac{3}{4}, red \frac{1}{4}
Bag B's pair goes after both bag A branches, because the bead from bag A does not change bag B.
Both white is one route, white then white: \frac{2}{5}\times\frac{3}{4}=\frac{6}{20}
Answer: \frac{6}{20}, which simplifies to \frac{3}{10}

Worked example 5: A fair spinner and a fair dice

Ravi has a fair 4-sided spinner numbered 1, 2, 3 and 4, and a fair 6-sided dice. He spins the spinner once and throws the dice once. Work out the probability that the spinner lands on 1 and the dice does not land on 6

The spinner's 4 numbers are equally likely: lands on 1 is \frac{1}{4}, not 1 is \frac{3}{4}
Your turn. The dice's 6 numbers are equally likely. What is the probability that the dice does not land on 6?

Your answer

5 of the 6 numbers are not 6: not 6 is \frac{5}{6}, and 6 is \frac{1}{6}
Keep quarters and sixths as fractions on the tree.
The route is "1" on the spinner, then "not 6" on the dice: \frac{1}{4}\times\frac{5}{6}=\frac{5}{24}
Answer: \frac{5}{24}

Worked example 6: What is wrong with this tree?

The probability that Nia scores when she throws a ball at a basket is 0.9. She takes two throws, and the first throw does not affect the second. Nia draws a probability tree diagram, but it is not correct. Write down what is wrong with it. Then work out the probability that Nia scores with both throws.

Here is Nia's tree. Her 1st throw pair is right: scores 0.9, misses 0.1
Her 2nd throw pairs add up to 0.45+0.45=0.9 and 0.05+0.05=0.1, not 1. She has shared out each 1st throw probability, as you would in a frequency tree.
The 2nd throw is just like the 1st, so both 2nd throw pairs should be scores 0.9, misses 0.1
Your turn. Work out the probability that Nia scores with both throws.

Your answer

Scores then scores: 0.9\times0.9=0.81
Answer: each 2nd throw pair should add up to 1 (scores 0.9, misses 0.1); the probability of scoring with both throws is 0.81

Common mistakes

Adding the probabilities along a route. For both events, multiply along the branches: 0.4\times0.5=0.2, not 0.4+0.5=0.9
Making a pair of branches add up to something other than 1. Every pair from the same point adds up to 1; the second pair does not share out the first branch's probability.
Mixing up the order on the second branches. Put each probability beside its own outcome, in the same order on every pair.
Losing a decimal place when multiplying. 0.2\times0.3=0.06, not 0.6: count the decimal places.
Changing thirds or sevenths into rounded decimals. Keep them as fractions so the answer stays exact.
Adding another route on to your answer. The question asks for one route: once you have multiplied along it, stop.