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Tree and Venn Diagrams

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.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.

Figure 3.10 Total = 64 + 24 + 24 + 9 = 121.

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 R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as follows:

R1R1, R1R2, R1R3, R2R1, R2R2, R2R3, R3R1, R3R2, R3R3.

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: B1R1, B1R2, B1R3, . . .

Show solution

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.

B1R1, B1R2, B1R3
B2R1, B2R2, B2R3
B3R1, B3R2, B3R3
B4R1, B4R2, B4R3
B5R1, B5R2, B5R3
B6R1, B6R2, B6R3
B7R1, B7R2, B7R3
B8R1, B8R2, B8R3

Problem

b. Calculate P(RR).

Show solution

Solution

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

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

Problem

c. Calculate P(RB OR BR).

Show solution

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) = $\frac{24 + 24}{121} = \frac{48}{121} \frac{24 + 24}{121} = \frac{48}{121}$ .

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) = $(\frac{3}{11}) (\frac{8}{11}) (\frac{3}{11}) (\frac{8}{11})$ + $(\frac{8}{11}) (\frac{3}{11}) (\frac{8}{11}) (\frac{3}{11})$ = $\frac{48}{121 .} \frac{48}{121 .}$

Problem

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

Show solution

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) = $\frac{24}{121} \frac{24}{121}$ .

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

Problem

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

Show solution

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.

Image shows a tree diagram with two levels the first level shows two branches. The first level represents the first draw of a marble. The left branch is labeled eight B and the right branch is labeled three R. The second level represents the second draw, two additional branches extend from the end of first level’s branches. The left branch of each is labeled eight B and the right branch is labeled three R

Figure 3.11

$P (R on 2nd draw GIVEN B on 1st) = P (R on 2nd | B on 1st) = \frac{3}{11} P (R) = P (R) = \frac{3}{11}$

You can also use the formula

P (R on 2nd | B on 1st) = \frac{P (R on 2nd AND B on 1st)}{P (B on 1st)} = \frac{\frac{24}{121}}{\frac{64 + 24}{121}} = \frac{24}{88} = \frac{3}{11 .} P (R) = \frac{P (R)}{P (B)} = \frac{\frac{24}{121}}{\frac{64 + 24}{121}} = \frac{24}{88} = \frac{3}{11 .}

Problem

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

Show solution

Solution

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

Problem

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

Show solution

Solution

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

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 $\frac{24}{33} \frac{24}{33}$ .

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).

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 12F and 40N. The second branch has a set of two lines (12F and 40N) for each line of the first branch. Multiply along each line to find 144FF, 480FN, 480NF, and 1,600NN.

Figure 3.12

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) = $(\frac{3}{11}) (\frac{2}{10}) = \frac{6}{110} (\frac{3}{11}) (\frac{2}{10}) = \frac{6}{110}$ .

This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.

Figure 3.13 Total =

\frac{56 + 24 + 24 + 6}{110} = \frac{110}{110} = 1 . \frac{56 + 24 + 24 + 6}{110} = \frac{110}{110} = 1 .

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) = ________

Show solution

Solution

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

Problem

b. Fill in the blanks.

P(RB OR BR) = $(\frac{3}{11}) (\frac{8}{10}) + (________)(________) = \frac{48}{110} (\frac{3}{11}) (\frac{8}{10}) + (________)(________) = \frac{48}{110}$

Show solution

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) = $(\frac{3}{11}) (\frac{8}{10}) (\frac{3}{11}) (\frac{8}{10})$ + $(\frac{8}{11}) (\frac{3}{10}) (\frac{8}{11}) (\frac{3}{10})$ = $\frac{48}{110} \frac{48}{110}$

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) = $\frac{3}{10} \frac{3}{10}$

Show solution

Solution

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

Problem

d. Fill in the blanks.

P(R on 1st AND B on 2nd) = P(RB) = (________)(________) = $\frac{24}{110} \frac{24}{110}$

Show solution

Solution

d. P(R on 1st AND B on 2nd) = P(RB) = $(\frac{3}{11}) (\frac{8}{10}) (\frac{3}{11}) (\frac{8}{10})$ = $\frac{24}{110} \frac{24}{110}$

Problem

e. Find P(BB).

Show solution

Solution

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

Problem

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

Show solution

Solution

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

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

This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).

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.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows 2 lines: F 12/52 and N 40/52. The second branch has a set of 2 lines (F 11/52 and N 40/51) for each line of the first branch. Multiply along each line to find FF 121/2652, FN 480/2652, NF 480/2652, and NN 1560/2652.

Figure 3.14

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.

This is a tree diagram with branches showing probabilities of kitten choices. The first branch shows two lines: T 4/9 and B 5/9. The second branch has a set of 2 lines for each first branch line. Below T 4/9 are T 3/8 and B 5/8. Below B 5/9 are T 4/8 and B 4/8. Multiply along each line to find probabilities of possible combinations.

Problem

Which shows the probability that both kittens are tabby?

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

$a. (\frac{4}{9}) (\frac{5}{9}) a. (\frac{4}{9}) (\frac{5}{9})$ $b. (\frac{4}{9}) (\frac{5}{8}) b. (\frac{4}{9}) (\frac{5}{8})$ $c. (\frac{4}{9}) (\frac{5}{9}) + (\frac{5}{9}) (\frac{4}{9}) c. (\frac{4}{9}) (\frac{5}{9}) + (\frac{5}{9}) (\frac{4}{9})$ $d. (\frac{4}{9}) (\frac{5}{8}) + (\frac{5}{9}) (\frac{4}{8}) d. (\frac{4}{9}) (\frac{5}{8}) + (\frac{5}{9}) (\frac{4}{8})$
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?

Show solution

Solution

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

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:

Image shows a Venn diagram consisting of two overlapping ovals inside a rectangle. The left oval is labeled A and the right is labeled B. The rectangle is labeled S. Annotations explain that Every outcome in the sample space is listed in the box. The shaded, overlapping area contains any outcome that appears in both events. This area contains six. All outcomes in A are listed in the oval labeled A. The values one, two, three, four, and five lie inside A, but outside the overlapping region. All outcomes in B are listed in the oval labeled B. The values seven, eight, and nine lie inside B, but outside the overlapping region. The outcomes ten, eleven, and twelve are in the sample space, but not in the events A or B, so these are listed in the region inside the rectangle and outside the ovals.

Figure 3.15

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:

This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.

Figure 3.16

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.

Image shows a Venn diagram consisting of two overlapping circles inside a rectangle. The left oval is labeled C and the right is labeled PT. The rectangle is labeled S. The overlapping region is labeled C and PT and contains the probability five hundredths. The region inside C, but outside the overlapping region, contains the probability thirty-five hundredths. The region inside B, but outside the overlapping region, contains the probability forty-five hundredths. The probability fifteen hundredths is listed in the region inside the rectangle and outside the ovals.

Figure 3.17

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.

Show solution

Solution

P(C) = .40

P(PT) = .50

P(C AND PT) = .05

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

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.

This is an empty Venn diagram showing two overlapping circles. The left circle is labeled O and the right circle is labeled RH-.

Figure 3.18

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.

Show solution

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.