7.04 Probability and tree diagrams

Tree diagrams

A tree diagram can be very useful to display the outcomes in multi-stage events, particularly if the number of options and stages is low.  At each point a tree diagram branches out to all the possible events that could occur from that point.

Trees and listing possible outcomes

When a single trial is carried out, we have just one column of branches and each outcome is listed at the end of each branch. Here are some examples:

 

Each column in the tree diagram represents a separate trial. In the diagram below, the first column represents the possible outcomes when a car encounters the first traffic light. The second column represents the possible outcomes the second time a car meets a traffic light. Reading the rows (branches) of the tree diagram from left to right we obtain the possible outcomes. So the first row gives us the outcome "first light green and second light green", the second row gives us the outcome "first light green and second light yellow". There are nine rows, therefore there are nine possible outcomes. 

Tree diagram showing first light encountered at two sets of traffic lights

Here are some more examples:

 

Probability from tree diagrams

 

Equally likely events

When outcomes are equally likely we can use our probability formula to determine the probability for an event.

$\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Probability=Number of favourable outcomesTotal number of outcomes​

Worked example

Example 1

A coin is tossed $3$3 times and the possible outcomes are presented in the tree diagram below:

(a) What is the probability of tossing all tails?

Think: There are $8$8 possible outcomes. How many result in the outcome $TTT$TTT?

Do:

$P\left(TTT\right)$P(TTT)

$=$=

$\frac{1}{8}$18​

 

(b) What is the probability of tossing exactly $2$2 tails?

Think: There are $8$8 possible outcomes. How many outcomes have exactly $2$2 tails? 

Do:

We would require one of the following outcomes: $TTH$TTH, $THT$THT or $HTT$HTT, hence:

$P\left(TTH,THT\text{ or }HTT\right)$P(TTH,THT or HTT)

$=$=

$\frac{3}{8}$38​

 

(c) What is the probability of tossing at least one head?

Think: There are $8$8 possible outcomes. How many outcomes have one or more heads? 

Do:

There are seven outcomes with at least $1$1 head, hence:

$P\left(\text{Number of heads}\ge1\right)$P(Number of heads≥1)

$=$=

$\frac{7}{8}$78​

 

Reflect: We could also calculate this as a complementary event. The outcomes that include at least one head is all the possible outcomes except all tails ($TTT$TTT). So we have:

$P\left(\text{Number of heads}\ge1\right)$P(Number of heads≥1)	$=$=	$1-P\left(TTT\right)$1−P(TTT)
	$=$=	$1-\frac{1}{8}$1−18​
	$=$=	$\frac{7}{8}$78​

 

Unequally likely events

When outcomes are not equally likely we can use a tree diagram to represent the situation by writing the probabilities on the branches. While we can also do this when outcomes are equally likely it is especially useful in cases where events have different weightings and can be used to calculate the probability in multi-stage events.

The important components of a weighted tree diagram are:

• Branch
• Probability 
• Outcome

 

 

 

Here are some examples that have probabilities on the branches, where the outcomes do not have an equal chance of occurring:

Notice that the sum of the branches from a single point always adds to $1$1 (or $100%$100%). This indicates that all the outcomes are listed.

To find the probability of an individual outcome multiply across the branches. Let's see how this works with an example.

Worked example

Example 2

The following probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is $\frac{3}{10}$310​ and the probability of losing is $\frac{7}{10}$710​.  (The probabilities can be written as fractions, decimals or percentages).

(a) Calculate the probability of the player winning both games.

$P\left(\text{WW}\right)$P(WW)	$=$=	$P\left(W\right)\times P\left(W\right)$P(W)×P(W)
	$=$=	$0.3\times0.3$0.3×0.3
	$=$=	$0.09$0.09

(b) Confirm the sum of the probabilities of all the outcomes is equal to $1$1.

$P\left(\text{WW or WL or LW or LL}\right)$P(WW or WL or LW or LL)	$=$=	$P\left(WW\right)+P\left(WL\right)+P\left(LW\right)+P\left(LL\right)$P(WW)+P(WL)+P(LW)+P(LL)
	$=$=	$0.09+0.21+0.21+0.49$0.09+0.21+0.21+0.49
	$=$=	$1$1

(c) Calculate the probability of the player winning only one game.

$P\left(\text{WL or LW}\right)$P(WL or LW)	$=$=	$P\left(WL\right)+P\left(LW\right)$P(WL)+P(LW)
	$=$=	$0.3\times0.7+0.7\times0.3$0.3×0.7+0.7×0.3
	$=$=	$0.42$0.42

(d) Calculate the probability of the player winning at least one game.

$P\left(\text{WW or WL or LW}\right)$P(WW or WL or LW)	$=$=	$P\left(WW\right)+P\left(WL\right)+P\left(LW\right)$P(WW)+P(WL)+P(LW)
	$=$=	$0.3\times0.3+0.3\times0.7+0.7\times0.3$0.3×0.3+0.3×0.7+0.7×0.3
	$=$=	$0.51$0.51

Or alternatively, use the complementary event of losing both games and calculate:

$P\left(\text{At least 1 win}\right)$P(At least 1 win)	$=$=	$1-P\left(LL\right)$1−P(LL)
	$=$=	$1-0.7\times0.7$1−0.7×0.7
	$=$=	$0.51$0.51

 

 

Tree diagrams for dependent events

For multistage events where the next stage is affected by the previous stage, we call these dependent events. We need to take care when drawing the tree diagram accordingly.

One type of experiment that is dependent on previous trials is an experiment without replacement. This means that the object selected (e.g. card, marble, person) is not able to be selected in any other selections. 

For example, the probability of drawing a red card from a standard pack of $52$52 cards is $\frac{26}{52}=\frac{1}{2}$2652​=12​. If we do draw a red card and choose to select a second card without replacement there are only $25$25 red cards left, but there are still $26$26 black cards. And there are only $51$51 cards left in the entire deck. The probability of selecting a second red card is $\frac{25}{51}$2551​. This can be seen in the top branches of the tree diagram above.

 

Practice questions

Question 3

Question 4

Question 5

Question 6