## What you need to know

### Probability Tree Diagrams

Probability is the study of how likely things are to happen. This topic will look at how probability tree’s, also known as tree diagrams, can be used to determine the probability of different types of events happening.

For basic probability concepts that feed into this topic view probability and frequency trees.

### Conditional Probability

If you are on the higher course, you also need to understand conditional probability – the chance of one thing happening given that you know something else has already happened. We use P(A|B) to denote “probability of B happening given A has happened”. If we were to ask, “What is the probability that B happens given that A has already happened?” then you can use a probability tree diagram effectively to answer this by going down the branch of A then looking at the probability of branch B occurring after this.

### Example 1: Making a Probability Tree

Ben flips an unbiased coin twice. Draw a tree diagram to represent the different outcomes of this event.

Each branch of the tree represents an event. Because the events in this example are not dependent on each other the probabilities do not change i.e. you have a 50:50 chance of heads or tails each time you flip the coin. We say these events are independent.

### Example 2: Using a Probability Tree

Penny will play 2 games of badminton against Monica. Penny’s chances of winning each game are outlined in the tree diagram below, where P denotes a Penny win and M denotes a Monica win. Work out the probability of Penny winning exactly one match.

In this example, the 2 matches are dependent – if Penny wins the first match, then the chance of her winning the second is 0.8, but if she loses the first match, her chance of winning the second is 0.5.

To use a tree diagram, we have to introduce the **and/or rule** of probability. On a tree diagram, we multiply along the branches, and then add the resulting probabilities at the end.

So, we are looking for the probability that Penny wins exactly one match. This means she could either win the first and lose the second or lose the first and win the second. By multiplying along the P line for the first match and the M line for the second, we get

\text{P(P wins 1st, M wins 2nd)}=0.7\times0.2=0.14.

Similarly, by multiplying along the M line for the first match and the P line for the second, we get

\text{P(M wins 1st, P wins 2nd)}=0.3\times0.5=0.15.

Either the first or second of these situations results in exactly one win for Penny, so to find the probability of one win for Penny, we add them: P(\text{Exactly 1 P win) }=0.14+0.15=0.29.

### Example 3: Constructing and Using a Probability Tree

A bag contains 4 red balls and 5 blue balls. Calculate the probability of selecting the same coloured ball each time, given that each time a ball is selected, it is not replaced.

Firstly, we need to construct the probability tree showing two selections, noting that there are 9 balls to begin with, reducing to 8 after the first selection, as shown below.

Now we need to use the tree diagram to determine the probability of selecting the same colour twice. We can see that there are two ways of doing this either blue and blue, or, red and red. Again we are using the and/or rule of probability via a tree diagram. This gives us

\text{P(blue and blue)}=\dfrac{5}{9}\times\dfrac{4}{8}=\dfrac{20}{73}

\text{P(red and red)}=\dfrac{4}{9}\times\dfrac{3}{8}=\dfrac{12}{73}

The final step then is to add the probabilities together to get

\text{P(Same colour)}=\dfrac{20}{73} +\dfrac{12}{73}=\dfrac{32}{73}

### Example Questions

**Question 1: **Anna and Rob take their driving tests on the same day. The probability of Anna passing her driving test is 0.7. The probability of both Anna and Rob passing is 0.35

**(a)** Work out the probability of Rob passing his driving test.

**(b)** Work out the probability of both Anna and Rob failing their driving tests.

**(a) **To work out the probability of Rob passing we can write the probability of both passing as:

P(A_p \cap R_p) =0.35

Substituting in the probability of Anna passing her test,

0.7 \times P(R_p) =0.35

Rearranging the equation to make P(R_p) the subject:

P(R_p) = 0.35 \div 0.7 = 0.5

**(b) ** The probability of both Anna and Rob failing their driving test can be found using a tree diagram as shown below:

Hence the probability of them both failing is **0.15 **

**Question 2: **There are 12 counters in a bag, 7 are blue and the rest are green.

Sean takes out a counter from the bag a random then, without replacement, takes out another counter.

Work out the probability that the two counters Sean removes are the same colour.

For this question when drawing the tree diagram we have to be careful as the probability changes between the two events. This is the result of not replacing the first counter hence only leaving 11 counters in the bag to pick from.

Adding together the probabilities of the result being blue then blue or green then green:

\dfrac{7}{22}+\dfrac{5}{33}=\dfrac{31}{66}

**Question 3: **The probability that a bus is on time is 0.75

Rory takes the bus to school two days a week . Calculate the probability the bus is late on each of those days.

To work out the probability of the bus being late on both of the days we can use a tree diagram where E represents the bus being on time or early and L represents the bus being late.

Going bottom line we find that the probability of being late of both days is:

\dfrac{1}{6}

**Question 4: **** **There are 14 footballs in a bag, 9 have a blue pattern design and the rest have green pattern design. The coach takes out a ball out from the bag a random then, without replacement, takes out another one.

Is the coach more likely to pick out two balls that are the same colour or two that are different colours.

You **must** show your workings. Give your answer in its simplest form.

Here we have to work out the probability that the coach takes out two balls that are a different colour.

For conditional probability questions, when drawing the tree diagram we have to be careful as the probability changes between the two events. This is the result of not replacing the first ball hence only leaving 13 balls in the bag to pick from.

Adding together the probabilities of the result being two different colours:

\dfrac{45}{182}+\dfrac{45}{182}=\dfrac{90}{182}=\dfrac{45}{91}

Hence as this is just below a half it is more likely that the coach picks two balls that are of the dame colour.

**Question 5: **William enters a badminton competition. The probability he wins a game is 0.6.

**(a)** Using this information complete the tree diagram shown below.

**(b) **Work out the probability that William wins at least one match.

**(a)** The resultant tree diagram should look something like:

**(b) **To find the probability he wins at least one game we can simple add the top 3 branches probabilities together or subtract the probability of bottom branch from 1:

\dfrac{9}{25}+\dfrac{6}{25}+\dfrac{6}{25}=\dfrac{21}{25}

or,

1-\dfrac{4}{25}=\dfrac{21}{25}

### Worksheets and Exam Questions

#### (NEW) Probability and Tree Diagrams Exam Style Questions - MME

Level 4-5#### Probability - Drill Questions

Level 1-3#### Probability And Relative Frequency - Drill Questions

Level 4-5#### Tree Diagrams - Drill Questions

Level 4-5#### Probability Trees - Drill Questions

Level 6-7#### Probability and Tree Diagrams - Drill Questions

Level 4-5### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.