10.03 Tree diagrams

A tree diagram is useful in tracking compound events. At each point it branches out to all the possible events that could occur from that point.

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

It is especially useful if the events have different weightings or are unequal events.

The important components of the tree diagram are:

• Branch

• Probability (used when the events have unequal probabilities)

• Outcome

If each outcome is equally likely the probability may be left off, if the outcomes are not equally likely then the probability will be written on the branches.

The sum of the probabilities for the branches from a single node should sum to 1.

When a single trial is carried out, we have just one column of branches.

Here are some examples. Neither of these have probabilities written on the branches because the outcomes are equally likely.

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

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

When more than one experiment is carried out, we have two (or more) columns of branches.

Here are some examples. These ones do not have the probabilities written, because the outcomes are equally likely.

Tree diagram for tossing a coin three times.

Tree diagram for whether a child gets home during the day or night during the next three days.

Here is an example that have probabilities on the branches. This probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is 0.3 and the probability of losing is 0.7.

The probabilities of the events are multiplied along each branch, for example the probability of winning both games is 9\% which is found by \\ 0.3 \times 0.3 = 0.09

To find the probability of at least 1 win, we could do either

a) \text{P(win, lose)} + \text{P(lose, win)} + \text{P(win, win)} = 21\% + 21\% + 9\% = 51\% or

b) Use the complementary event of losing both games and calculate:

1 - \text{P (lose, lose)} = 1- 49\% = 51\%

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 cards is \dfrac{26}{52} = \dfrac{1}{2}. If we do draw a red card and choose to select a second card without replacement there are only 25 red cards left, but there are still 26 black cards. And there are only 51 cards left in the entire deck. The probability of selecting a second red card is \dfrac{25}{51}. This can be seen in the top branches of the tree diagram.

Examples

Example 1

Three fair coins are tossed.

a

Find the probability of obtaining at least one head.

Worked Solution

Create a strategy

Count the number of times one head or more occurs and the total number of possible outcomes, and then write them as a fraction.

Apply the idea

The heads occurs on 7 occassions. The total number of possible outcomes is 8.

P(H)=\dfrac{7}{8}

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b

What is the probability of obtaining TTH in this sequence?

Worked Solution

Create a strategy

Count the number of times TTH occurs and the total number of possible outcomes, and then write them as a fraction.

Apply the idea

There is one occurence of TTH. The total number of possible outcomes is 8.

P(TTH)=\dfrac{1}{8}

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c

What is the probability of obtaining THH in this sequence?

Worked Solution

Create a strategy

Count the number of times THH occurs and the total number of possible outcomes, and then write them as a fraction.

Apply the idea

There is one occurence of THH. The total number of possible outcomes is 8.

P(THH)=\dfrac{1}{8}

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Example 2

The proportion of scholarship recipients at a particular university is \dfrac{7}{10}. The number of students at the university is so large that even if a student is removed, we can say that the proportion of scholarship recipients remains the same. If three students are selected at random:

a

What is the probability that at least one of the students is a scholarship recipient?

Worked Solution

Create a strategy

We can use the formula: P(R)=1-P(N).

Apply the idea

\displaystyle P(R)	\displaystyle =	\displaystyle 1-P(N)	Write the formula
\displaystyle P(R)	\displaystyle =	\displaystyle 1-\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}	Substitute the N=\dfrac{3}{10}
	\displaystyle =	\displaystyle 1-\dfrac{27}{1000}	Evaluate the multiplication
	\displaystyle =	\displaystyle \dfrac{973}{1000}	Evaluate the subtraction

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b

What is the probability that at least one of the students is a nonrecipient?

Worked Solution

Create a strategy

We can use the formula: P(N) = 1 - P(R).

Apply the idea

\displaystyle P(N)	\displaystyle =	\displaystyle 1-P(R)	Write the formula
\displaystyle P(N)	\displaystyle =	\displaystyle 1-\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}	Substitute the R=\dfrac{7}{10}
	\displaystyle =	\displaystyle 1-\dfrac{343}{1000}	Evaluate the multiplication
	\displaystyle =	\displaystyle \dfrac{657}{1000}	Evaluate the subtraction

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c

What is the probability there is at least one recipient and one nonrecipient in the selection?

Worked Solution

Create a strategy

Multiply the probabilities along the branches of the tree that include at least one recipient and at least one nonrecipient, then sum the individual probabilities.

Apply the idea

Disregard the branches where all the selected students are recipients and where all the selected students are nonrecipients.

\displaystyle \text{Probability}	\displaystyle =	\displaystyle (P(R) \times P(R) \times P(N)) + (P(R) \times P(N) \times P(R))+ \\ (P(N) \times P(R) \times P(R)) + (P(N) \times P(N) \times P(R)) + \\ (P(N) \times P(R) \times P(N)) + (P(R) \times P(N) \times P(N))	Write the formula
	\displaystyle =	\displaystyle \left(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\right) + \left(\dfrac{7}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\right) + \\ \left(\dfrac{3}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}\right) + \left(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\right) + \\ \left(\dfrac{3}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\right) + \left(\dfrac{7}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}\right)	Substitute \\R=\dfrac{7}{10},\,N=\dfrac{3}{10}
	\displaystyle =	\displaystyle \dfrac{147}{1000}+\dfrac{147}{1000}+\dfrac{147}{1000}+ \dfrac{63}{1000}+\dfrac{63}{1000}+\dfrac{63}{1000}	Evaluate the multiplication
	\displaystyle =	\displaystyle \dfrac{63}{100}	Evaluate the addition

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