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7.04 Probability and tree diagrams

Lesson

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:

The outcomes for tossing a coin once Outcomes from rolling a standard die

 

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:

 

Tree diagram for tossing a coin three times Tree diagram for selecting two cards from a deck and noting the colour.

 

For the first example given above we have $8$8 possible outcomes and reading across the branches we can see the outcomes are: $\left\{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\right\}${TTT,TTH,THT,THH,HTT,HTH,HHT,HHH} and for the second example of drawing two cards and noting the colour we have $4$4 outcomes where are $\left\{RR,RB,BR,BB\right\}${RR,RB,BR,BB}.

 

Practice questions

Question 1

Construct a tree diagram showing all possible outcomes of boys and girls a couple with three children can possibly have.

question 2

Construct a tree diagram showing all the ways a captain and a vice-captain can be selected from Matt, Rebecca and Helen.

 

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 heads1) $=$= $\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 heads1) $=$= $1-P\left(TTT\right)$1P(TTT)
  $=$= $1-\frac{1}{8}$118
  $=$= $\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:

Outcomes of selecting a face card or number card from deck of cards

Outcomes of certain person playing a game of chess (win, lose or draw)

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

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)$1P(LL)
  $=$= $1-0.7\times0.7$10.7×0.7
  $=$= $0.51$0.51

 

Probabilities from tree diagrams
  • Multiply along the branches to calculate the probability of individual outcomes.
  • Add down the list of outcomes to calculate the probability of multiple options.
  • The sum of the final probabilities for all the outcomes should be $1$1$\left(100%\right)$(100%) - this is useful to check if you have calculated everything correctly.

 

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

Three cards labelled $\editable{2}$2, $\editable{3}$3 and $\editable{4}$4 are placed face down on a table. Two of the cards are selected randomly to form a two-digit number. The outcomes are displayed in the following probability tree.

  1. List the sample space of two digit numbers produced by this process.

    Separate different outcomes with a comma.

  2. Find the probability that $2$2 appears as a digit in the number.

  3. Find the probability that the sum of the two selected cards is even.

  4. What is the probability of forming a number greater than $40$40?

Question 4

Every morning Mae has toast for breakfast. Each day she either chooses honey or jam to spread on her toast, with equal chance of choosing either one.

  1. Draw a tree diagram for three consecutive days of Mae’s breakfast choices.

  2. What is the probability that on the fourth day Mae chooses honey for her toast?

  3. What is the probability that Mae chooses jam for her toast three days in a row?

Question 5

Han plays three tennis matches. In each match he has $60%$60% chance of winning:

  1. Find the probability that he will win all his matches.

  2. Find the probability that he will lose all his matches.

  3. Find the probability that he will win more matches than he loses.

Question 6

Bart is purchasing a plane ticket to Adelaide. He notices there are only $4$4 seats remaining, $1$1 of them is a window seat (W) and the other $3$3 are aisle seats (A). His friend gets on the computer and purchases a ticket immediately after. The seats are randomly allocated at the time of purchase.

  1. Fill in the probabilities matching the edges of the probability tree for the seat Bart receives and the seat his friend receives:

     $\editable{}$  $\editable{}$  $\editable{}$
     $\editable{}$    $\editable{}$
  2. What is the probability that Bart's friend has an aisle seat?

  3. What is the probability of Bart's friend receiving an aisle seat if Bart has a window seat?

Outcomes

4.1.8

use a sample space to determine the probability of outcomes for an experiment

4.1.9

use arrays or tree diagrams to determine the outcomes and the probabilities for experiments

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