Tree Diagram

Let's take a football tournament, with play extending so that the only two possibilities are win or lose. In the first match, a team has a 60% chance of winning. If they win in the first match, the chances of winning the second game extend to 80%, whereas if they lose, it decreases to a 40% chance of winning. Show this information in a tree diagram.

First, we will denote a win by W and a loss by L. The first event is the first match.

Step 1: There are two events, so we need to draw two lines.

Step 2: We will denote one of these lines with a W at the end and the other with an L. This looks like the below.

Depiction of a tree diagram event - Tom Maloy, StudySmarter Originals

Step 3: If there is a 60% chance of winning, this means there is a 40% chance of losing, as the two options must sum to 100%. In terms of decimals, this means we have a 0.6 chance of winning and a 0.4 chance of losing. We can now add this to the diagram. (If you need a refresher on this, revise your knowledge on converting decimals and percentages)

Tree diagram event with probabilities - Tom Maloy, StudySmarter Originals

Step 4: We now need to repeat this process for the next branches. As there are again two outcomes in the second event, we draw two branches off of each branch, and then we label these W and L to represent winning and losing.

The probability of winning after already winning is 0.8, so the probability of losing after a win is 0.2. The probability of winning after a loss is 0.4, so the probability of losing the second match in a row is 0.6. We can now fill these probabilities in on our tree diagram.

Chain of tree diagram events with probabilities - Tom Maloy, StudySmarter Originals

Following on from the above example, find the probability of a team winning one match and losing another, in any order.

The first thing we are going to do is multiply along each branch, to get the probability of each outcome occurring. The results of this are below.

Tree diagram example - Tom Maloy, StudySmarter Originals

If we want one win and one loss, then the team can lose the first game and win the second, or win the first and lose the second. That means we need to add together P(W, L) and P(L, W), which gets us 0.12+0.16=0.28.

Example 1:

I have ten balls in a bag; five are green, three are yellow, and two are blue. I take one ball out of the bag and do not replace it. I then take another ball.

1. Draw a tree diagram to represent this scenario

2. Find the probability of taking two balls of different colours.

3. What is the probability of choosing two balls, neither of which are yellow?

a) Let us first find the probability of each ball in the first ball draw. For green, we have $5/10$, for yellow we have $3/10$, and for blue, we have $2/10=1/5$. We can display this information on a tree diagram, where we use B to represent blue, Y for yellow and G for green.

Tree diagram for question a) - Tom Maloy, StudySmarter Originals

When we have taken one green ball out, we have nine balls total left, with four green, three yellow and two blue, so the probability of choosing green is $4/9$, choosing yellow has probability $3/9$and blue has probability $2/9$.

When we have taken one yellow ball out, we have nine balls total left, with five green, two yellow and two blue, so the probability of choosing green is $5/9$, choosing yellow has probability $2/9$and blue has probability $2/9$.

When we have taken one blue ball out, we have nine balls total left, with five green, three yellow and one blue, so the probability of choosing green is $5/9$, choosing yellow has probability $3/9$ and blue has probability $1/9$. This is shown in the tree diagram below.

Tree diagram for question a) - Tom Maloy, StudySmarter Originals

We will now multiply through the branches to get the probabilities of each possibility.

Tree diagram for question a) - Tom Maloy, StudySmarter Originals

b) For two balls of different colours, we need to add the various branches. This gives us $$\begin{gathered} P(B,Y)+P(Y,B)+P(B,G)+P(G,B)+P(Y,G)+P(G,Y) \\ =15/90+10/90+15/90+6/90+10/90+6/90=62/90=31/45 \end{gathered}$$

c) For two balls, neither yellow, we again add branches. We get $$\begin{gathered} P(G,G)+P(B,B)+P(G,B)+P(B,G) \\ =20/90+2/90+10/90+10/90=42/90=7/15 \end{gathered}$$

Example 2:

Below is a tree diagram. Fill in the gaps, and then use it to find the probability of two R and one B and the probability of getting the same letter three times.

Tree diagram for question b) -Tom Maloy, StudySmarter Originals

In each pair of corresponding branches, the probability must sum to one. Where there is 0.7 in one branch, the corresponding branch must be marked by 0.3. The same goes for 0.4 with 0.6, 0.2 with 0.8 and 0.1 with 0.9. Filling these in, we get the result below. Once we have done this, we can multiply along each branch to show the probability of that branch. This is also shown on the diagram below.

Tree diagram for question b) -Tom Maloy, StudySmarter Originals

To get two R's and a B, we can go RRB, RBR or BRR, so we need to sum these probabilities together. P(RRB)+P(RBR)+P(BRR)=0.224+0.294+0.036=0.554

To get the same letter three times, we can either have BBB or RRR. Adding the probabilities results in 0.056+0.021=0.077.