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3.4: Contingency Tables and Probability Trees

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    4558
  • Contingency Tables

    A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.

    Example 3.20

    Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

    Speeding violation in the last yearNo speeding violation in the last yearTotal
    Uses cell phone while driving25280305
    Does not use cell phone while driving45405450
    Total70685755

    Table3.2

    The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

    Calculate the following probabilities using the table.

    a. Find P(Driver is a cell phone user).

    Answer

    Solution 3.20

    a. \(\frac{\text { number of cell phone users }}{\text { total number in study }}=\frac{305}{755}\)

    b. Find P(Driver had no violation in the last year).

    Answer

    Solution 3.20

    b. \(\frac{\text { number that had no violation }}{\text { total number in study }}=\frac{685}{755}\)

    c. Find P(Driver had no violation in the last year \(\cap\) was a cell phone user).

    Answer

    Solution 3.20

    c. \(\frac{280}{755}\)

    d. Find P(Driver is a cell phone user \(\cup\) driver had no violation in the last year).

    Answer

    Solution 3.20

    d. \(\left(\frac{305}{755}+\frac{685}{755}\right)-\frac{280}{755}=\frac{710}{755}\)

    e. Find P(Driver is a cell phone user \(|\) driver had a violation in the last year).

    Answer

    Solution 3.20

    e. \(\frac{25}{70}\) (The sample space is reduced to the number of drivers who had a violation.)

    f. Find P(Driver had no violation last year \(|\) driver was not a cell phone user)

    Answer

    Solution 3.20

    f. \(\frac{405}{450}\) (The sample space is reduced to the number of drivers who were not cell phone users.)

    Exercise 3.20

    Table 3.3 shows the number of athletes who stretch before exercising and how many had injuries within the past year.

    Injury in last yearNo injury in last yearTotal
    Stretches55295350
    Does not stretch231219450
    Total286514800

    Table3.3

    1. What is P(athlete stretches before exercising)?
    2. What is P(athlete stretches before exercising||no injury in the last year)?

      Example 3.21

      Table 3.4 shows a random sample of 100 hikers and the areas of hiking they prefer.

      SexThe coastlineNear lakes and streamsOn mountain peaksTotal
      Female1816___45
      Male______1455
      Total___41______

      Table3.4 Hiking Area Preference

      a. Complete the table.

      Answer

      Solution 3.21

      a.

      SexThe coastlineNear lakes and streamsOn mountain peaksTotal
      Female18161145
      Male16251455
      Total344125100

      Table3.5 Hiking Area Preference

      b. Are the events "being female" and "preferring the coastline" independent events?

      Let F = being female and let C = preferring the coastline.

      1. Find \(P(F\cap C)\).
      2. Find P(F)P(C)

      Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.

      Answer

      Solution 3.21

      b.

      1. \(P(F\cap C)=\frac{18}{100}\) = 0.18
      2. P(F)P(C) = \(\left(\frac{45}{100}\right)\left(\frac{34}{100}\right)\) = (0.45)(0.34) = 0.153

      \(P(F\cap C)\) ≠ P(F)P(C), so the events F and C are not independent.

      c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.

      1. What word tells you this is a conditional?
      2. Fill in the blanks and calculate the probability: P(___||___) = ___.
      3. Is the sample space for this problem all 100 hikers? If not, what is it?
      Answer

      Solution 3.21

      c.

      1.The word 'given' tells you that this is a conditional.

      2.P(M||L) = \(\frac{25}{41}\)

      3.No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

      d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P= prefers mountain peaks.

      1. Find P(F).
      2. Find P(P).
      3. Find \(P(F\cap P)\).
      4. Find \(P(F\cup P)\).
      Answer

      Solution 3.21

      d.

      1. P(F) = \(\frac{45}{100}\)
      2. P(P) = \(\frac{25}{100}\)
      3. \(P(F\cap P)\)= \(\frac{11}{100}\)
      4. \(P(F\cup P)\)= \(\frac{45}{100}+\frac{25}{100}-\frac{11}{100}=\frac{59}{100}\)

        Exercise 3.21

        Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.

        GenderLake pathHilly pathWooded pathTotal
        Female453827110
        Male26521290
        Total719039200

        Table3.6

        1. Out of the males, what is the probability that the cyclist prefers a hilly path?
        2. Are the events “being male” and “preferring the hilly path” independent events?

          Example 3.22

          Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1515 and the probability he is not caught is 4545. If he goes out the second door, the probability he gets caught by Alissa is 1414 and the probability he is not caught is 3434. The probability that Alissa catches Muddy coming out of the third door is 1212 and the probability she does not catch Muddy is 1212. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1313.

          Caught or notDoor oneDoor twoDoor threeTotal
          Caught\(\frac{1}{15}\)\(\frac{1}{12}\)\(\frac{1}{6}\)____
          Not caught\(\frac{4}{15}\)\(\frac{3}{12}\)\(\frac{1}{6}\)____
          Total____________1

          Table3.7 Door Choice

          • The first entry \(\frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right)\) is \(P(Door One\cap Caught)\)
          • The entry \(\frac{4}{15}=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)\) is \(P(Door One\cap Not Caught)\)

          Verify the remaining entries.

          a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

          Answer

          Solution 3.22

          a.

          Caught or notDoor oneDoor twoDoor threeTotal
          Caught\(\frac{1}{15}\)\(\frac{1}{12}\)\(\frac{1}{6}\)\(\frac{19}{60}\)
          Not caught\(\frac{4}{15}\)\(\frac{3}{12}\)\(\frac{1}{6}\)

          \(\frac{41}{60}\)

          Total\(\frac{5}{15}\)\(\frac{4}{12}\)\(\frac{2}{6}\)1

          Table3.8 Door Choice

          b. What is the probability that Alissa does not catch Muddy?

          Answer

          Solution 3.22

          b. \(\frac{41}{60}\)

          c. What is the probability that Muddy chooses Door One \cap Door Two given that Muddy is caught by Alissa?

          Answer

          Solution 3.22

          c. \(\frac{9}{19}\)

          Example 3.23

          Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

          YearRobberyBurglaryRapeVehicleTotal
          2008145.7732.129.7314.7
          2009133.1717.729.1259.2
          2010119.370127.7239.1
          2011113.7702.226.8229.6
          Total

          Table3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

          TOTAL each column and each row. Total data = 4,520.7

          1. Find \(P(2009\cap Robbery)\).
          2. Find \(P(2010\cap Burglary)\).
          3. Find \(P(2010\cup Burglary)\).
          4. Find P(2011|Rape).
          5. Find P(Vehicle|2008).
          Answer

          Solution 3.23

          1. 0.0294
          2. 0.1551
          3. 0.7165
          4. 0.2365
          5. 0.2575

          Exercise 3.23

          Table 3.10 relates the weights and heights of a group of individuals participating in an observational study.

          Weight/heightTallMediumShortTotals
          Obese182814
          Normal205128
          Underweight12259
          Totals

          Table3.10

          1. Find the total for each row and column
          2. Find the probability that a randomly chosen individual from this group is Tall.
          3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
          4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
          5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
          6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
          7. Are the events Obese and Tall independent?

            Tree Diagrams

            Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams 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. 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.

            Figure 3.2 Total = 64 + 24 + 24 + 9 = 121

            The first set of branches represents the first draw. The second set of branches represents the second draw. 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:

            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.

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

            Answer

            Solution 3.24

            a. 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. Using the tree diagram, calculate P(RR).

            Answer

            Solution 3.24

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

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

            Answer

            Solution 3.24

            c. \(P(RB\cup BR)\) = \(\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)=\frac{48}{121}\)

            d. Using the tree diagram, calculate \(P(Ron 1st draw\cap Bon 2nd draw)\).

            Answer

            Solution 3.24

            d. \(P(Ron 1st draw\cap Bon 2nd draw) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right)=\frac{24}{121}\)

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

            Answer

            Solution 3.24

            e. P(R on 2nd draw|B on 1st draw) = P(R on 2nd|B on 1st) = \(\frac{24}{88} = \frac{3}{11}\)

            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. \(\frac{24}{88} = \frac{3}{11}\).

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

            Answer

            Solution 3.24

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

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

            Answer

            Solution 3.24

            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. 24 of the 33 outcomes have B on the second draw. The probability is then \(\frac{24}{33}\).

            Exercise 3.24

            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.

            Figure 3.3

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

            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 3.4 Total = \(\frac{56+24+24+6}{110}=\frac{110}{110}=1\)

            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 ten marbles left in the urn.

            Calculate the following probabilities using the tree diagram.

            a. P(RR) = ________

            Answer

            Solution 3.25

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

            b. Fill in the blanks:

            \(P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\) + (___)(___) = \(\frac{48}{110}\)

            Answer

            Solution 3.25

            b. \(P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)=\frac{48}{110}\)

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

            Answer

            Solution 3.25

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

            d. Fill in the blanks.

            \(P(Ron 1st\cap Bon 2nd)\) = (___)(___) = \(\frac{24}{100}\)

            Answer

            Solution 3.25

            d. \(P(Ron 1st\cap Bon 2nd) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)=\frac{24}{100}\)

            e. Find P(BB).

            Answer

            Solution 3.25

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

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

            Answer

            Solution 3.25

            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.

            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)

            Exercise 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 3.5

            1. Find \(P(FN\cup NF)\).
            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.

              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.
              1. What is the probability that both kittens are tabby?

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

                a.\(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)\) b.\(\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)\) c.\(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)\) 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?
              Answer

              Solution 3.26

              a. c, b. d, c. \(\frac{4}{8}\), d. \(\frac{32}{72}\)

                Exercise 3.26

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