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Chapter 3: Probability Topics

3.5 Tree and Venn Diagrams

Learning Objectives

By the end of this section, you should be able to:

  • Use tree diagrams to find probabilities for experiments with and without replacement
  • Use Venn diagrams to visualize relationships between events

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

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, which means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

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.

The first set of branches represents the first draw, where there are 8 ways to first draw B and three ways to first draw R. The second set of branches represents the second draw, with the same potential outcomes for each of the original branches due to the replacement of the first draw. Each of the outcomes is distinct. In fact, we can label 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:

{ R1R1, R1R2, R1R3, R2R1, R2R2, R2R3, R3R1, R3R2, R3R3}. Counting these shows 9 ways we can draw two red balls. We can also get this total from our tree diagram by multiplying down the branches: 3 times 3 = 9 RR outcomes. The other branches are similar; there are 64 ways to draw two blue balls, 24 ways to draw blue then red, and 24 ways to draw red then blue, for a total of 64+24+24+9 = 121 outcomes of this experiment.

We can also find this total by multiplying: there are a total of 11 balls in the urn. Then the experiment to draw two balls, one at a time, with replacement has (11)(11) = 121 outcomes, the size of the sample space.

We can now use our tree diagram to compute some probabilities.

  1. Calculate [latex]P(RR)[/latex].
  2. Calculate [latex]P(RB \text{ OR } BR).[/latex]
  3. Calculate [latex]P(\text{R on 1st draw AND B on 2nd draw})[/latex].
  4. Calculate [latex]P(R\text{ 2nd } | B \text{ 1st} ).[/latex]
  5. Calculate [latex]P(BB).[/latex]
  6. Calculate [latex]P(B\text{ 2nd } | R \text{ 1st} ).[/latex]

1. Using the tree diagram, calculate [latex]P(RR)[/latex].

Solution

[latex]P(RR) = \left(\frac{3}{11}\right)\left(\frac{3}{11}\right) = \frac{9}{121}[/latex]

 

2. Using the tree diagram, calculate [latex]P(RB \text{ OR } BR)[/latex].

Solution

[latex]P(RR \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{11}\right) = \frac{48}{121}[/latex]

 

3. Using the tree diagram, calculate [latex]P(\text{R on 1st draw AND B on 2nd draw})[/latex].

Solution

P(R on 1st draw AND B on 2nd draw) = P(RB) = [latex]\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)[/latex] = [latex]\frac{24}{121}[/latex]

 

4. Using the tree diagram, calculate [latex]P(\text{R on 2nd draw GIVEN B on 1st draw})[/latex].

Solution

[latex]P(\text{R on 2nd draw GIVEN B on 1st draw}) = P(\text{R on 2nd|B on 1st}) = \frac{24}{88} = \frac{3}{11}[/latex]

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are [latex]24 + 64 = 88[/latex] possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. [latex]\frac{24}{88}[/latex] = [latex]\frac{3}{11}[/latex].

 

5. Using the tree diagram, calculate [latex]P(BB)[/latex].

Solution

[latex]P(BB) = \frac{64}{121}[/latex]

 

6. Using the tree diagram, calculate [latex]P(\text{B on the 2nd draw GIVEN R on the first draw})[/latex].

Solution

[latex]P(\text{B on 2nd draw|R on 1st draw}) = \frac{8}{11}[/latex]

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then [latex]9 + 24 = 33[/latex]. 24 of the 33 outcomes have B on the second draw. The probability is then [latex]\frac{24}{33}[/latex].

Your Turn!

In a standard deck, there are 52 cards. 12 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.
Solution

Total number of outcomes is [latex]144 + 480 + 480 + 1600 = 2,704[/latex].

[latex]P(FF) = \frac{144}{\text{144 + 480 + 480 + 1,600}}=\frac{144}{2,704}=\frac{9}{169}[/latex]

Your Turn!

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, [latex]\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}[/latex].

Total = [latex]\frac{56+24+24+6}{110}=\frac{110}{110}=1[/latex]

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.

Note: Because we are drawing without replacement, on the second draw there are ten marbles left in the urn because we do not put back (replace) the first marble. This means that on the second draw, we have different numbers of red and blue marbles remaining in the two branches. If you draw a red on the first draw from the three red possibilities, there are two red marbles and eight blue marbles left to draw on the second draw.

Calculate the following probabilities using the tree diagram.

a. [latex]P(RR) = \underline{\hspace{2cm}}[/latex]

Solution

[latex]P(RR) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}[/latex]

 

b. Fill in the blanks:

[latex]P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + (\underline{\hspace{2cm}})(\underline{\hspace{2cm}}) = \frac{48}{110}[/latex]

Solution

[latex]P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{48}{110}[/latex]

 

c. [latex]P(\text{R on 2nd|B on 1st}) = \underline{\hspace{2cm}}[/latex]

Solution

[latex]P(\text{R on 2nd|B on 1st}) = \frac{3}{10}[/latex]

 

d. Fill in the blanks.

[latex]P(\text{R on 1st AND B on 2nd}) = P(RB) = (\underline{\hspace{2cm}})(\underline{\hspace{2cm}}) = \frac{24}{100}[/latex]

Solution

[latex]P(\text{R on 1st AND B on 2nd}) = P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{100}[/latex]

 

e. Find [latex]P(BB)[/latex].

Solution

[latex]P(BB) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right)[/latex]

 

f. Find [latex]P(\text{B on 2nd|R on 1st})[/latex].

Solution

Using the tree diagram, [latex]P(\text{B on 2nd|R on 1st}) = P(R|B) = \frac{8}{10}[/latex].

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).

  • [latex]P(R|R)[/latex] here means [latex]P(\text{R on 2nd|R on 1st})[/latex].
  • [latex]P(B|R)[/latex] here means [latex]P(\text{B on 2nd|R on 1st})[/latex].
  • [latex]P(R|B)[/latex] here means [latex]P(\text{R on 2nd|B on 1st})[/latex].
  • [latex]P(B|B)[/latex] here means [latex]P(\text{B on 2nd|B on 1st})[/latex].

 

Your Turn!

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.
  1. Find [latex]P(FN \text{ OR } NF)[/latex].
  2. Find [latex]P(N|F)[/latex].
  3. Find [latex]P(\text{at most one face card})[/latex].
    Hint: “At most one face card” means zero or one face card.
  4. Find [latex]P(\text{at least on face card})[/latex].
    Hint: “At least one face card” means one or two face cards.
Solution
  1. [latex]P(FN \text{ OR } NF) = \frac{480}{2,652} + \frac{480}{2,652} = \frac{960}{2,652} = \frac{80}{221}[/latex]
  2. [latex]P(N|F) = \frac{40}{51}[/latex]
  3. [latex]P(\text{at most one face card}) = \frac{(480 + 480 + 1,560)}{2,652}[/latex] = [latex]\frac{2,520}{2,652}[/latex]
  4. [latex]P(\text{at least one face card})\frac{(132 + 480 + 480)}{2,652}[/latex] = [latex]\frac{1,092}{2,652}[/latex]

Your Turn!

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.

  1. What is the probability that both kittens are tabby?
    a.[latex]\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)[/latex]
    b.[latex]\left(\frac{4}{9}\right)\left(\frac{4}{9}\right)[/latex]
    c.[latex]\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)[/latex]
    d.[latex]\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)[/latex]
  2. What is the probability that one kitten of each coloring is selected?
    a.[latex]\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)[/latex]
    b.[latex]\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)[/latex]
    c.[latex]\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)[/latex]
    d.[latex]\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)[/latex]
  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?
Solution

1. c
2. d
3. [latex]\frac{4}{8}[/latex]
4. [latex]\frac{32}{72}[/latex]

Venn Diagrams

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

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:

A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.

Your Turn!

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.

Solution

Example

Forty percent of the students at a local college belong to a club and 50% 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.

This is a venn diagram with one set containing students in clubs and another set containing students working part-time. Both sets share students who are members of clubs and also work part-time. The universe is labeled S.

If a student is selected at random, find

  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.
Solution
  1. [latex]P(C) = 0.40[/latex]
  2. [latex]P(PT) = 0.50[/latex]
  3. [latex]P(C \text{ AND } PT) = 0.05[/latex]
  4. [latex]P(C|PT) = \frac{P(C\text{ AND }PT)}{P(PT)} = \frac{0.05}{0.50} = 0.1[/latex]
  5. [latex]P(C \text{ OR } PT) = P(C) + P(PT) - P(C \text{ AND } PT) = 0.40 + 0.50 - 0.05 = 0.85[/latex]

 

Your Turn!

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% of African Americans have the Rh- factor, and 51% 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-.

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

We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.

  1. [latex]P(O) = \underline{\hspace{2cm}}[/latex]
  2. [latex]P(R) = \underline{\hspace{2cm}}[/latex]
  3. [latex]P(O \text{ AND } R) = \underline{\hspace{2cm}}[/latex]
  4. [latex]P(O \text{ OR } R) = \underline{\hspace{2cm}}[/latex]
  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.
Solution
  1. 0.51
  2. 0.075
  3. 0.04
  4. 0.545
  5. The area represents the African Americans that have type O blood and the Rh- factor.
  6. The area represents the African Americans that have neither type O blood nor the Rh- factor.

Your Turn!

In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a nonfiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.

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

1. and 4. In the following Venn diagram below, the blue oval represents customers buying a novel, the red oval represents customers buying nonfiction, and the yellow oval represents customers who buy compact disks.

2. [latex]P(\text{novel or nonfiction}) = P(\text{Blue OR Red}) = P(\text{Blue}) + P(\text{Red}) - P(\text{Blue AND Red}) = 0.6 + 0.4 - 0.2 = 0.8[/latex]

3. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book.

Section 3.5 Review

A tree diagram uses branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.

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. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.

Section 3.5 Practice

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 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation.

Solution
This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.

Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, [latex]P(H) = \frac{2}{3}[/latex] and P(T) = [latex]\frac{1}{3}[/latex] where H is heads and T is tails.

Tree diagram with 2 branches. The first branch consists of 2 lines of H=2/3 and T=1/3. The second branch consists of 2 sets of 3 lines each with the both sets containing R=3/12, Y=4/12, and B=5/12.

1. Find [latex]P(\text{tossing a Head on the coin AND a Red bead})[/latex]

a. [latex]\frac{2}{3}[/latex]
b. [latex]\frac{5}{15}[/latex]
c. [latex]\frac{6}{36}[/latex]
d. [latex]\frac{5}{36}[/latex]

 

2. Find [latex]P(\text{Blue bead})[/latex].

a. [latex]\frac{15}{36}[/latex]
b. [latex]\frac{10}{36}[/latex]
c. [latex]\frac{10}{12}[/latex]
d. [latex]\frac{6}{36}[/latex]

Solution

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 [latex]P(S)[/latex].
  5. Let T be the event that the cookies selected were different flavors. Find [latex]P(T)[/latex] 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 [latex]P(U)[/latex].

Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.

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 [latex]P(G_1 \text{ AND } G_2)[/latex].
  3. Find [latex]P(\text{at least one green})[/latex].
  4. Find [latex]P(G_2|G_1)[/latex].
  5. Are G2 and G1 independent events? Explain why or why not.
Solution
  1. 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. [latex]P(GG) = \left(\frac{5}{8}\right)\left(\frac{5}{8}\right)[/latex] = [latex]\frac{25}{64}[/latex]
  3. [latex]P(\text{at least one green}) = P(GG) + P(GY) + P(YG) = \frac{25}{64}[/latex] + [latex]\frac{15}{64}[/latex] + [latex]\frac{15}{64}[/latex] = [latex]\frac{55}{64}[/latex]
  4. [latex]P(G|G) = \frac{5}{8}[/latex]
  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 [latex]P(G_1 \text{ AND } G_2)[/latex].
  3. Find [latex]P(\text{at least one green})[/latex].
  4. Find [latex]P(G_2|G_1)[/latex].
  5. Are G2 and G1 independent events? Explain why or why not.

Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.

Complete the following.

  1. Construct a table or a tree diagram of the situation.
  2. Find [latex]P(\text{driver is female})[/latex].
  3. Find [latex]P(\text{driver is age 65 or over|driver is female})[/latex].
  4. Find [latex]P(\text{driver is age 65 or over AND female})[/latex].
  5. In words, explain the difference between the probabilities in part 3 and part 4.
  6. Find [latex]P(\text{driver is age 65 or over})[/latex].
  7. Are being age 65 or over and being female mutually exclusive events? How do you know?
Solution
  1. <20 20–64 >64 Totals
    Female 0.0244 0.3954 0.0661 0.486
    Male 0.0259 0.4186 0.0695 0.514
    Totals 0.0503 0.8140 0.1356 1
  2. [latex]P(F) = 0.486[/latex]
  3. [latex]P(>64|F) = 0.1361[/latex]
  4. [latex]P(>64 \text{ AND } F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661[/latex]
  5. [latex]P(>64|F)[/latex] 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. [latex]P(>64) = P(>64 \text{ AND } F) + P(>64 \text{ AND } M) = 0.1356[/latex]
  7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time [latex]P(>64 \text{AND } F) = 0.0661[/latex].

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% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% 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?
Solution
  1. Car, Truck or Van Walk Public Transportation Other Totals
    Alone 0.7318
    Not Alone 0.1332
    Totals 0.8650 0.0390 0.0530 0.0430 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: [latex]P(\text{Alone}) = 0.7318 + 0.0390 = 0.7708[/latex].
  3. Make the same assumptions as in part 2 we have: [latex](0.7708)(1,000) = 771[/latex]
  4. [latex](0.1332)(1,000) = 133[/latex]

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 [latex]P(H)[/latex] and [latex]P(T)[/latex].
  2. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
  3. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
  4. Use the tree to find the probability of obtaining at least one head.

The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories:

* includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 ____
Male 2,146 463 60 135 ____
Totals ____ ____ ____ ____ ____

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

  1. Find [latex]P(\text{Person is female})[/latex].
  2. Find [latex]P(\text{Person has a risk factor heterosexual contact})[/latex].
  3. Find [latex]P(\text{Person is female OR has a risk factor of IV drug user})[/latex].
  4. Find [latex]P(\text{Person is female AND has a risk factor of homosexual/bisexual})[/latex].
  5. Find [latex]P(\text{Person is male AND has a risk factor of IV drug user})[/latex].
  6. Find [latex]P(\text{Person is female GIVEN person got the disease from heterosexual contact})[/latex].
  7. Construct a Venn diagram. Make one group females and the other group heterosexual contact.
Solution

The completed contingency table is as follows:

* includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 255
Male 2,146 463 60 135 2,804
Totals 2,146 533 196 184 3,059
  1. [latex]\frac{255}{3059}[/latex]
  2. [latex]\frac{196}{3059}[/latex]
  3. [latex]\frac{718}{3059}[/latex]
  4. [latex]0[/latex]
  5. [latex]\frac{463}{3059}[/latex]
  6. [latex]\frac{136}{196}[/latex]

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

  1. Find [latex]P(\text{Person is female})[/latex].
  2. Find [latex]P(\text{Person obtained the disease through heterosexual contact})[/latex].
  3. Find [latex]P(\text{Person is female GIVEN person got the disease from heterosexual contact})[/latex].
  4. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.

References

Data from Clara County Public H.D.

Data from the American Cancer Society.

Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013).

Data from the Federal Highway Administration, part of the United States Department of Transportation.

Data from the United States Census Bureau, part of the United States Department of Commerce.

Data from USA Today.

“Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013).

“Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html (accessed May 2, 2013).

 

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