Tree and Venn Diagrams

a. List the 24 BR outcomes: B1R1, B1R2, B1R3, . . .

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

b. Calculate P(RR).

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) = $\left(\frac{3}{11}\right)\left(\frac{3}{11}\right)$ = $\frac{9}{121\text{.}}$

c. Calculate P(RB OR BR).

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

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

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) = $\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)$ = $\frac{24}{121}$

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

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.

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

You can also use the formula

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

f. P(BB) = $\frac{64}{121}$

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

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 $\frac{24}{33}$.

a. P(RR) = ________

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

b. Fill in the blanks.

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

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

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

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

d. Fill in the blanks.

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

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

e. Find P(BB).

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

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

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

1. Which shows the probability that both kittens are tabby?
   $\text{a.}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$ $\text{b.}\left(\frac{4}{9}\right)\left(\frac{4}{9}\right)$ $\text{c.}\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)$ $\text{d.}\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$
2. What is the probability that one kitten of each coloring is selected?
   $\text{a.}\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$ $\text{b.}\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)$ $\text{c.}\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)$ $\text{d.}\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)$
3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
4. What is the probability of choosing two kittens of the same color?

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

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:

1. the probability that the student belongs to a club.
2. the probability that the student works part time.
3. the probability that the student belongs to a club AND works part time.
4. the probability that the student belongs to a club given that the student works part time.
5. the probability that the student belongs to a club OR works part time.

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

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.

1. P(O) = ___________
2. P(R) = ___________
3. P(O AND R) = ___________
4. P(O OR R) = ____________
5. In the Venn Diagram, describe the overlapping area using a complete sentence.
6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

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.

The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false-positive test result, meaning the test comes back for cancer when the man does not have it, is .51. Let C = a man develops cancer in his lifetime; P = a man has at least one false-positive test. Construct a tree diagram of the situation.

Find P(tossing a head on the coin AND a red bead).

1. $\frac{2}{3}$
2. $\frac{5}{15}$
3. $\frac{6}{36}$
4. $\frac{5}{36}$

Find P(blue bead).

1. $\frac{15}{36}$
2. $\frac{10}{36}$
3. $\frac{10}{12}$
4. $\frac{6}{36}$

a

A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. How many cookies did he take?

1. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
2. Are the probabilities for the flavor of the second cookie that Miguel selects independent of his first selection? Explain.
3. For each complete path through the tree, write the event it represents and find the probabilities.
4. Let S be the event that both cookies selected were the same flavor. Find P(S).
5. Let T be the event that the cookies selected were different flavors. Find P(T) by two different methods by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
6. Let U be the event that the second cookie selected is a butter cookie. Find P(U).

Suppose that you randomly draw two cards, one at a time, with replacement.
Let G₁ = first card is green
Let G₂ = second card is green

1. Draw a tree diagram of the situation.
2. Find P(G₁ AND G₂).
3. Find P(at least one green).
4. Find P(G₂|G₁).
5. Are G₂ and G₁ independent events? Explain why or why not.

1. View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.

2. P(GG) = $\left(\frac{5}{8}\right)\left(\frac{5}{8}\right)$ = $\frac{25}{64}$
3. P(at least one green) = P(GG) + P(GY) + P(YG) = $\frac{25}{64}$ + $\frac{15}{64}$ + $\frac{15}{64}$ = $\frac{55}{64}$
4. P(G|G) = $\frac{5}{8}$
5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn. The composition of cards in the bag remains the same from draw one to draw two.

Suppose that you randomly draw two cards, one at a time, without replacement.
G₁ = first card is green
G₂ = second card is green

1. Draw a tree diagram of the situation.
2. Find P(G₁ AND G₂).
3. Find P(at least one green).
4. Find P(G₂|G₁).
5. Are G₂ and G₁ independent events? Explain why or why not.

Complete the following:

1. Construct a table or a tree diagram of the situation.
2. Find P(driver is female).
3. Find P(driver is age 65 or over|driver is female).
4. Find P(driver is age 65 or over AND female).
5. In words, explain the difference between the probabilities in Part c and Part d.
6. Find P(driver is age 65 or over).
7. Are being age 65 or over and being female mutually exclusive events? How do you know?

1. 	<20	20–64	>64	Totals
   Female	.0244	.3954	.0661	.486
   Male	.0259	.4186	.0695	.514
   Totals	.0503	.8140	.1356	1

2. P(F) = .486
3. P(>64|F) = .1361
4. P(>64 and F) = P(F) P(>64|F) = (.486)(.1361) = .0661
5. P(>64|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older.
6. P(>64) = P(>64 and F) + P(>64 and M) = .1356
7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) = .0661.

Suppose that 10,000 U.S. licensed drivers are randomly selected.

1. How many would you expect to be male?
2. Using the table or tree diagram, construct a contingency table of gender versus age group.
3. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.

Approximately 86.5 percent of Americans commute to work by car, truck, or van. Out of that group, 84.6 percent drive alone and 15.4 percent drive in a carpool. Approximately 3.9 percent walk to work and approximately 5.3 percent take public transportation.

1. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
2. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
3. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work?
4. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool?

1. 	Car, Truck or Van	Walk	Public Transportation	Other	Totals
   Alone	.7318
   Not Alone	.1332
   Totals	.8650	.0390	.0530	.0430	1

2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = .7318 + .0390 = .7708.
3. Make the same assumptions as in (b) we have: (.7708)(1,000) = 771
4. (.1332)(1,000) = 133

When the euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one euro-coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin.

1. Based on the given data, find P(H) and P(T).
2. Use a tree to find the probabilities of each possible outcome for the experiment of spinning the coin twice.
3. Use the tree to find the probability of obtaining exactly one head in two spins of the coin.
4. Use the tree to find the probability of obtaining at least one head.

Use the following information to answer the next two exercises. The following are real data from Santa Clara County, California. As of a certain time, there had been a total of 3,059 documented cases of a disease in the county. They were grouped into the following categories, with risk factors of becoming ill with the disease labeled as Methods A, B, and C and Other:

Suppose a person with a disease in Santa Clara County is randomly selected.

1. Find P(Person is female).
2. Find P(Person has a risk factor of method C).
3. Find P(Person is female OR has a risk factor of method B).
4. Find P(Person is female AND has a risk factor of method A).
5. Find P(Person is male AND has a risk factor of method B).
6. Find P(Person is female GIVEN person got the disease from method C).
7. Construct a Venn diagram. Make one group females and the other group method C.

The completed contingency table is as follows:

1. $\frac{255}{3059}$
2. $\frac{196}{3059}$
3. $\frac{718}{3059}$
4. 0
5. $\frac{463}{3059}$
6. $\frac{136}{196}$

7. View original figure (external; image not reproduced in this text edition)

   Original figure description: Two ovals are positioned next to each other horizontally, with a small overlap. The left oval is labeled F, contains the number 119, and is purple. The right oval is labeled HC, contains the number 60, and is yellow. The space in between contains the number 136 and is light blue.

Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of a disease had been reported in Santa Clara County, California, through a certain date. Those cases will be our population. Of those cases, 6.4 percent obtained the disease through method C and 7.4 percent are female. Out of the females with the disease, 53.3 percent got the disease from method C.

1. Find P(Person is female).
2. Find P(Person obtained the disease through method C).
3. Find P(Person is female GIVEN person got the disease from method C)
4. Construct a Venn diagram representing this situation. Make one group females and the other group method C. Fill in all values as probabilities.