Probability and Tree Diagrams Worksheets, Questions and Revision

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}