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(RB).[/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]
Solution

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

c. Using the tree diagram, calculate P(RB OR BR).

Solution

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

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

Solution

d. 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]

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

Solution

e. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = [latex]\frac{24}{88}[/latex]
= [latex]\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 24 + 64 = 88 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].

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

Solution

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

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) = [latex]\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 9 + 24 = 33. 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 144 + 480 + 480 + 1600 = 2,704.

P(FF) = [latex]\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. P(RR) = ________

Solution

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

b. Fill in the blanks:

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

Solution

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

c. P(R on 2nd|B on 1st) =

Solution

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

d. Fill in the blanks.

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

Solution

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

e. Find P(BB).

Solution

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

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) = [latex]\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).

  • 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)

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

    Hint: “At most one face card” means zero or one face card.

  4. Find P(at least on face card).

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

Solution
  1. P(FN OR NF) = [latex]\frac{\text{480}}{\text{2,652}}\text{ + }\frac{\text{480}}{\text{2,652}}\text{ = }\frac{\text{960}}{\text{2,652}}\text{ = }\frac{\text{80}}{\text{221}}[/latex]
  2. P(N|F) = [latex]\frac{40}{51}[/latex]
  3. P(at most one face card) = [latex]\frac{\text{(480 + 480 + 1,560)}}{\text{2,652}}[/latex] = [latex]\frac{2,520}{2,652}[/latex]
  4. P(at least one face card) = [latex]\frac{\text{(132 + 480 + 480)}}{\text{2,652}}[/latex] = [latex]\frac{\text{1,092}}{\text{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

a. c, b. d, c. [latex]\frac{4}{8}[/latex], d. [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

  • the probability that the student belongs to a club. P(C) = 0.40
  • the probability that the student works part time. P(PT) = 0.50
  • the probability that the student belongs to a club AND works part time. P(C AND PT) = 0.05
  • the probability that the student belongs to a club given that the student works part time. [latex]P\text{(}C\text{|}PT\text{)} = \frac{P\text{(}C\text{ AND }PT\text{)}}{P\text{(}PT\text{)}} = \frac{0.05}{0.50} = 0.1[/latex]
  • the probability that the student belongs to a club OR works part time. P(C OR PT) = P(C) + P(PT) – P(C AND PT) = 0.40 + 0.50 – 0.05 = 0.85

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. 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.
Solution

a. 0.51; b. 0.075; c. 0.04; d. 0.545; 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.

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

a. and d. 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.

b. P(novel or nonfiction) = P(Blue OR Red) = P(Blue) + P(Red) – P(Blue AND Red) = 0.6 + 0.4 – 0.2 = 0.8.

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

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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