Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.

## Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of *branches* that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram:

## Example 3.24

In an urn, there are 11 balls. Three balls are red (*R*) and eight balls are blue (*B*). Draw two balls, one at a time, **with replacement**. *With replacement* means that you put the first ball back in the urn before you select the second ball. Therefore, you are selecting from exactly the same group each time, so each draw is independent. The tree diagram shows all the possible outcomes.

The first set of branches represents the first draw. There are 8 ways to draw a blue marble and 3 ways to draw a red one. The second set of branches represents the second draw. Regardless of the choice on the first draw, there are again eight ways to draw a blue marble and 3 ways to draw a red one. Read down each branch to see the total number of possible outcomes. For example, there are 8 ways to get a blue marble on the first draw, and eight ways to get one on the second draw, so there are 8 × 8 = 64 different ways to draw two blue marbles in succession. Each of the outcomes is distinct. In fact, we can list each red ball as *R*1, *R*2, and *R*3 and each blue ball as *B*1, *B*2, *B*3, *B*4, *B*5, *B*6, *B*7, and *B*8. Then the nine *RR* outcomes can be written as follows:

*R*1*R*1, *R*1*R*2, *R*1*R*3, *R*2*R*1, *R*2*R*2, *R*2*R*3, *R*3*R*1, *R*3*R*2, *R*3*R*3.

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

### Problem

a. List the 24 *BR* outcomes: *B*1*R*1, *B*1*R*2, *B*1*R*3, . . .

### Solution

a. We know that there will be 24 different possible outcomes because there are eight ways to draw blue and three ways to draw red. Make a systematic list of possible outcomes that consist of a blue marble on the first draw and a red marble on the second draw.

*B*1*R*1, *B*1*R*2, *B*1*R*3

*B*2*R*1, *B*2*R*2, *B*2*R*3

*B*3*R*1, *B*3*R*2, *B*3*R*3

*B*4*R*1, *B*4*R*2, *B*4*R*3

*B*5*R*1, *B*5*R*2, *B*5*R*3

*B*6*R*1, *B*6*R*2, *B*6*R*3

*B*7*R*1, *B*7*R*2, *B*7*R*3

*B*8*R*1, *B*8*R*2, *B*8*R*3

### Problem

b. Calculate *P*(*RR*).

### Solution

b. You can use the tree diagram. There are nine ways to draw two reds and 121 possible outcomes. So, *P*(*RR*) = $\frac{\text{9}}{\text{121}\text{.}}$.

Each draw is independent, so you can also use the formula: *P*(*RR*) = *P*(*R*)*P*(*R*) = $\left(\frac{3}{11}\right)\left(\frac{3}{11}\right)$ = $\frac{9}{121\text{.}}$

### Problem

c. Calculate *P*(*RB* OR *BR*).

### Solution

c. The tree diagram shows that there are 24 ways to draw *RB* and 24 ways to draw *BR*. There are 121 possible outcomes, so *P*(*RB* or *BR*) = $\frac{24+24}{121}=\frac{48}{121}$.

The events *RB* and *BR* are mutually exclusive, so *P*(*RB* OR *BR*) = *P*(*RB*) + *P*(*BR*) = *P*(*R*)*P*(*B*) + *P*(*B*)*P*(*R*) = $\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)$ + $\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)$ = $\frac{48}{121\text{.}}$

### Problem

d. Using the tree diagram, calculate *P*(*R* on 1st draw AND *B* on 2nd draw).

### Solution

d. Follow the path on the tree. There are three ways to get a red marble on the first draw and eight ways to get a blue on the second draw. There are 3 × 8 = 24 ways to draw red then blue, so *P*(*RB*) = $\frac{24}{121}$.

Can you think of another way to find this probability? *P*(*R* on 1st draw AND *B* on 2nd draw) = *P*(*RB*) = $\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)$ = $\frac{24}{121}$

### Problem

e. Using the tree diagram, calculate *P*(*R* on 2nd draw GIVEN *B* on 1st draw).

### Solution

e. Given that a blue marble is selected first, we need only follow the left set of branches on the tree diagram. In this case, there are three ways to obtain red on the second draw and 11 possible outcomes.

$$P\left(R\phantom{\rule{.2em}{0ex}}\text{on 2nd draw GIVEN}\phantom{\rule{.2em}{0ex}}B\phantom{\rule{.2em}{0ex}}\text{on 1st}\right)=P\left(R\phantom{\rule{.2em}{0ex}}\text{on 2nd |}\phantom{\rule{.2em}{0ex}}B\phantom{\rule{.2em}{0ex}}\text{on 1st}\right)=\frac{3}{11}$$

You can also use the formula

### Problem

f. Using the tree diagram, calculate *P*(*BB*).

### Solution

f. *P*(*BB*) = $\frac{64}{121}$

### Problem

g. Using the tree diagram, calculate *P*(*B* on the 2nd draw GIVEN *R* on the first draw).

### Solution

g. *P*(*B* on 2nd draw|*R* on 1st draw) = $\frac{8}{11}$

There are 9 + 24 outcomes that have *R* on the first draw (9 *RR* and 24 *RB*). The sample space is then 9 + 24 = 33. Twenty-four of the 33 outcomes have *B* on the second draw. The probability is then $\frac{24}{33}$.

## Try It 3.24

In a standard deck, there are 52 cards. Twelve cards are face cards (event *F*) and 40 cards are not face cards (event *N*). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate *P*(*FF*).

## Example 3.25

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. ** Without replacement** means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example,

*P*(

*RR*) = $\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$.

## NOTE

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw **without replacement**, so that on the second draw there are 10 marbles left in the urn.

Calculate the following probabilities using the tree diagram:

### Problem

a. *P*(*RR*) = ________

### Solution

a. *P*(*RR*) = $\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$

### Problem

b. Fill in the blanks.

*P*(*RB* OR *BR*) = $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\text{}+\text{(\_\_\_\_\_\_\_\_)(\_\_\_\_\_\_\_\_)}=\text{}\frac{48}{110}$

### Solution

b. *P*(*RB* OR *BR*) = *P*(*RB*) + *P*(*BR*) = *P*(*R* on 1st) *P*(*B* on 2nd) + *P*(*B* on 1st) *P*(*R* on 2nd) = $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$ + $\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)$ = $\frac{48}{110}$

### Problem

c. Because this is a conditional probability, we restrict the sample space to consider only those outcomes that have a blue marble in the first draw. Look at the second level of the tree to see that *P*(*R* on 2nd|*B* on 1st) = $\frac{3}{10}$

### Solution

c. *P*(*R* on 2nd|*B* on 1st) = $\frac{3}{10}$

### Problem

d. Fill in the blanks.

*P*(*R* on 1st AND *B* on 2nd) = *P*(*RB*) = (________)(________) = $\frac{24}{100}$

### Solution

d. *P*(*R* on 1st AND *B* on 2nd) = *P*(*RB*) = $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$ = $\frac{24}{100}$

### Problem

e. Find *P*(*BB*).

### Solution

e. *P*(*BB*) = $\left(\frac{8}{11}\right)\left(\frac{7}{10}\right)$

### Problem

f. Find *P*(*B* on 2nd|*R* on 1st).

### Solution

f. Using the tree diagram, *P*(*B* on 2nd|*R* on 1st) = *P*(*R*|*B*) = $\frac{8}{10}$.

If we are using probabilities, we can label the tree in the following general way:

*P*(*R*|*R*) here means*P*(*R*on 2nd|*R*on 1st)*P*(*B*|*R*) here means*P*(*B*on 2nd|*R*on 1st)*P*(*R*|*B*) here means*P*(*R*on 2nd|*B*on 1st)*P*(*B*|*B*) here means*P*(*B*on 2nd|*B*on 1st)

## Try It 3.25

In a standard deck, there are 52 cards. Twelve cards are face cards (*F*) and 40 cards are not face cards (*N*). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

- Find
*P*(*FN*OR*NF*). - Find
*P*(*N*|*F*). - Find
*P*(at most one face card).

Hint:*At most one face card*means zero or one face card. - Find
*P*(at least one face card).

Hint:*At least one face card*means one or two face cards.

## Example 3.26

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

### Problem

- Which shows the probability that both kittens are tabby?

$\text{a.}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$ $\text{b.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{4}{9}}\right)$ $\text{c.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{3}{8}}\right)$ $\text{d.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)$ - What is the probability that one kitten of each coloring is selected?

$\text{a.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)$ $\text{b.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{8}}\right)$ $\text{c.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)+\left({\scriptscriptstyle \frac{5}{9}}\right)\left({\scriptscriptstyle \frac{4}{9}}\right)$ $\text{d.}\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{8}}\right)+\left({\scriptscriptstyle \frac{5}{9}}\right)\left({\scriptscriptstyle \frac{4}{8}}\right)$ - What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
- What is the probability of choosing two kittens of the same color?

### Solution

a. $\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{3}{8}}\right)$, b. $\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{8}}\right)+\left({\scriptscriptstyle \frac{5}{9}}\right)\left({\scriptscriptstyle \frac{4}{8}}\right)$, c. $\frac{4}{8}$, d. $\frac{32}{72}$

## Try It 3.26

Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

## Venn Diagram

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space *S* together with circles or ovals. The circles or ovals represent events.

## Example 3.27

Suppose an experiment has the outcomes 1, 2, 3, . . . , 12 where each outcome has an equal chance of occurring. Let event *A* = {1, 2, 3, 4, 5, 6} and event *B* = {6, 7, 8, 9}. Then *A* AND *B* = {6} and *A* OR *B* = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

## Try It 3.27

Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event *C* = {green, blue, purple} and event *P* = {red, yellow, blue}. Then *C* AND *P* = {blue} and *C* OR *P* = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.

## Example 3.28

Flip two fair coins. Let *A* = tails on the first coin. Let *B* = tails on the second coin. Then *A* = {*TT*, *TH*} and *B* = {*TT*, *HT*}. Therefore, *A* AND *B* = {*TT*}. *A* OR *B* = {*TH*, *TT*, *HT*}.

The sample space when you flip two fair coins is *X* = {*HH*, *HT*, *TH*, *TT*}. The outcome *HH* is in NEITHER *A* NOR *B*. The Venn diagram is as follows:

## Try It 3.28

Roll a fair, six-sided die. Let *A* = a prime number of dots is rolled. Let *B* = an odd number of dots is rolled. Then *A* = {2, 3, 5} and *B* = {1, 3, 5}. Therefore, *A* AND *B* = {3, 5}. *A* OR *B* = {1, 2, 3, 5}. The sample space for rolling a fair die is *S* = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

## Example 3.29

### Problem

**Forty percent** of the students at a local college belong to a club and **50 percent** work part time. **Five percent** of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let *C* = student belongs to a club and *PT* = student works part time.

Start by drawing a rectangle to represent the sample space. Then draw two circles or ovals inside the rectangle to represent the events of interest: belonging to a club (*C*) and working part time (*PT*). Always draw overlapping shapes to represent outcomes that are in both events.

Label each piece of the diagram clearly and note the probability or frequency of each part. Start by labeling the overlapping section first. Note that the probabilities in *C* total 0.40 and the sum of the probabilities in *PT* is 0.50. The total of all probabilities displayed must be 1, representing 100 percent of the sample space.

If a student is selected at random, find the following:

- the probability that the student belongs to a club.
- the probability that the student works part time.
- the probability that the student belongs to a club AND works part time.
- the probability that the student belongs to a club
**given**that the student works part time. - the probability that the student belongs to a club
**OR**works part time.

### Solution

*P*(*C*) = .40

*P*(*PT*) = .50

*P*(*C* AND *PT*) = .05

$P\text{(}C\text{|}PT\text{)}=\frac{P\text{(}C\text{AND}PT\text{)}}{P\text{(}PT\text{)}}=\frac{.05}{.50}=.1$

*P*(*C* OR *PT*) = *P*(*C*) + *P*(*PT*) − *P*(*C* AND *PT*) = .40 + .50 − .05 = .85

## Try It 3.29

Fifty percent of the workers at a factory work a second job, 25 percent have a spouse who also works, and 5 percent work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let *W* = works a second job and *S* = spouse also works.

## Example 3.30

### Problem

A person with type O blood and a negative Rh factor (Rh–) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative Rh factor, 5−10 percent of African Americans have the Rh– factor, and 51 percent have type O blood.

The “O” circle represents the African Americans with type O blood. The “Rh––" oval represents the African Americans with the Rh– –factor.

We will use the average of 5 percent and 10 percent, 7.5 percent, as the percentage of African Americans who have the Rh–– factor. Let *O* = African American with Type O blood and *R* = African American with Rh– –factor.

*P*(*O*) = ___________*P*(*R*) = ___________*P*(*O*AND*R*) = ___________*P*(*O*OR*R*) = ____________- In the Venn Diagram, describe the overlapping area using a complete sentence.
- In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

### Solution

a. *P*(*O*) = .51

b. *P*(*R*) = .075 because an average of 7.5 percent of African Americans have the Rh– –factor.

c. *P*(*O* AND *R*) = 0.04 because 4 percent of African Americans have both Type O blood and the Rh– –factor.

d. *P*(*O* OR *R*) = *P*(*O*) + *P*(*R*) - *P*(*O* AND *R*) = .51 + .075 − .04 = .545

e. The area represents the African Americans that have type O blood and the Rh–– factor.

f. The area represents the African Americans that have neither type O blood nor the Rh–– factor.

## Try It 3.30

In a bookstore, the probability that the customer buys a novel is .6, and the probability that the customer buys a nonfiction book is .4. Suppose that the probability that the customer buys both is .2.

- Draw a Venn diagram representing the situation.
- Find the probability that the customer buys either a novel or a nonfiction book.
- In the Venn diagram, describe the overlapping area using a complete sentence.
- Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.