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3.14: Chapter 3 Solution (Practice + Homework)

  • Page ID
    5549
  • 1.

    1. \(P(L′) = P(S)\)
    2. \(P(M \cup S)\)
    3. \(P(F \cap L)\)
    4. \(P(M|L)\)
    5. \(P(L|M)\)
    6. \(P(S|F)\)
    7. \(P(F|L)\)
    8. \(P(F \cup L)\)
    9. \(P(M \cap S)\)
    10. \(P(F)\)

    3.

    \(P(N)=\frac{15}{42}=\frac{5}{14}=0.36\)

    5.

    \(P(C)=\frac{5}{42}=0.12\)

    7.

    \(P(G)=\frac{20}{150}=\frac{2}{15}=0.13\)

    9.

    \(P(R)=\frac{22}{150}=\frac{11}{75}=0.15\)

    11.

    \(P(O)=\frac{150-22-38-20-28-26}{150}=\frac{16}{150}=\frac{8}{75}=0.11\)

    13.

    \(P(E)=\frac{47}{194}=0.24\)

    15.

    \(P(N)=\frac{23}{194}=0.12\)

    17.

    \(P(S)=\frac{12}{194}=\frac{6}{97}=0.06\)

    19.

    \(\frac{13}{52}=\frac{1}{4}=0.25\)

    21.

    \(\frac{3}{6}=\frac{1}{2}=0.5\)

    23.

    \(P(R)=\frac{4}{8}=0.5\)

    25.

    \(P(O \cup H)\)

    27.

    \(P(H|I)\)

    29.

    \(P(N|O)\)

    31.

    \(P(I \cup N)\)

    33.

    \(P(I)\)

    35.

    The likelihood that an event will occur given that another event has already occurred.

    37.

    1

    39.

    the probability of landing on an even number or a multiple of three

    41.

    \(P(J) = 0.3\)

    43.

    \(P(Q\cap R)=P(Q)P(R)\)

    \(0.1 = (0.4)P(R)\)

    \(P(R) = 0.25\)

    45.

    0.376

    47.

    C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.

    49.

    L \cap C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.

    51.

    0.6492

    53.

    No, because P(L \cap C) does not equal 0.

    55.

    \(P(\text { musician is a male } \cap \text { had private instruction) }=\frac{15}{130}=\frac{3}{26}=0.12.\)

    57.

    \(P(\text { being a female musician } \cap \text { learning music in school) }=\frac{38}{130}=\frac{19}{65} = 0.29\)

    \(P(\text { being a female musician }) P(\text { learning music in school })=\left(\frac{72}{130}\right)\left(\frac{62}{130}\right)=\frac{4,464}{16,900}=\frac{1,116}{4,225}=0.26\)

    No, they are not independent because \(P(\text { being a female musician } \cap \text { learning music in school) }\) is not equal to \(P( \text{being a female musician})P(\text {learning music in school})\).

    58.

    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'.

    Figure 3.21

    60.

    \(\frac{35,065}{100,450}\)

    62.

    To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is \(\frac{4,715}{100,450}\).

    64.

    To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is \(\frac{4715}{15,273}\).

    66.

    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.

      Figure 3.22

    2. \(P(G G)=\left(\frac{5}{8}\right)\left(\frac{5}{8}\right)=\frac{25}{64}\)
    3. \(P(\text { at least one green })=P(G G)+P(G Y)+P(Y G)=\frac{25}{64}+\frac{15}{64}+\frac{15}{64}=\frac{55}{64}\)
    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.

    68.

    1. <20>20–64>64Totals
      Female0.02440.395464">0.06610.486
      Male0.02590.418664">0.06950.514
      Totals0.05030.814064">0.13561

      Table3.22

    2. \(P(F) = 0.486\)
    3. \(P(>64 | F) = 0.1361\)
    4. \(P(>64 \text{ and } F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661\)
    5. \(P(>64 | F)\) is the percentage of female drivers who are 65 or older and P(>64 \cap F) is the percentage of drivers who are female and 65 or older.
    6. \(P(>64) = P(>64 \cap F) + P(>64 \cap M) = 0.1356\)
    7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time \(P(>64 \cap F) = 0.0661\).

    70.

    1. Car, truck or vanWalkPublic transportationOtherTotals
      Alone0.7318
      Not alone0.1332
      Totals0.86500.03900.05300.04301

      Table3.23

    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: \(P(\text{Alone}) = 0.7318 + 0.0390 = 0.7708\).
    3. Make the same assumptions as in (b) we have: \((0.7708)(1,000) = 771\)
    4. \((0.1332)(1,000) = 133\)

    73.

    1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
    2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.

    75.

    0

    77.

    0.3571

    79.

    0.2142

    81.

    Physician (83.7)

    83.

    \(83.7 − 79.6 = 4.1\)

    85.

    \(P(\text{Occupation} < 81.3) = 0.5\)

    87.

    1. The Forum Research surveyed 1,046 Torontonians.
    2. 58%
    3. 42% of 1,046 = 439 (rounding to the nearest integer)
    4. 0.57
    5. 0.60.

    89.

    1. \(P(\text { Betting on two line that touch each other on the table) }=\frac{6}{38}.\)
    2. \(P(\text { Betting on three numbers in a line })=\frac{3}{38}\)
    3. \(P(\text { Betting on one number })=\frac{1}{38}\)
    4. \(P(\text { Betting on four number that touch each other to form a square) }=\frac{4}{38}.\)
    5. \(P(\text { Betting on two number that touch each other on the table })=\frac{2}{38}\)
    6. \(P(\text { Betting on } 0-00-1-2-3)=\frac{5}{38}\)
    7. \(P(\text { Betting on } 0-1-2 ; \text { or } 0-00-2 ; \text { or } 00-2-3)=\frac{3}{38}\)

    91.

    1. \(\{G1, G2, G3, G4, G5, Y1, Y2, Y3\}\)
    2. \(\frac{5}{8}\)
    3. \(\frac{2}{3}\)
    4. \(\frac{2}{8}\)
    5. \(\frac{6}{8}\)
    6. No, because \(P(G \cap E)\) does not equal 0.

    93.

    NOTE

    The coin toss is independent of the card picked first.

    1. \(\{(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)\}\)
    2. \(P(A)=P(\text { blue }) P(\text { head })=\left(\frac{3}{10}\right)\left(\frac{1}{2}\right)=\frac{3}{20}\)
    3. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). \(P(A \cap B) = 0\)
    4. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). \(P(A \cap C) = P(A) = \frac{3}{20}\)

    95.

    1. \(S = \{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)\}\)
    2. \(\frac{4}{8}\)
    3. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, \(P(A \cap B) = 0\).

    97.

    1. If Y and Z are independent, then \(P(Y \cap Z) = P(Y)P(Z)\), so \(P(Y \cup Z) = P(Y) + P(Z) - P(Y)P(Z)\).
    2. 0.5

    99.

    iii i iv ii

    101.

    1. \(P(R) = 0.44\)
    2. \(P(R|E) = 0.56\)
    3. \(P(R|O) = 0.31\)
    4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; \(P(R|E) \neq P(R)\).
    5. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; \(P(R|E) > P(R)\).

    103.

    1. \(P(\text { type } \mathrm{O} \cup \mathrm{Rh}-)=P(\text { type } \mathrm{O})+P(\mathrm{Rh}-)-P(\text { type } \mathrm{O} \cap \mathrm{Rh}-)\)

      \(0.52=0.43+0.15-P(\text { type } O \cap \mathrm{Rh}-)\); solve to find \(P(\text { type } \mathrm{O} \cap \mathrm{Rh}-)= 0.06\)

      6% of people have type O, Rh- blood

    2. \(P(\text { NOT (type O } \cap \mathrm{Rh}-) )=1-P(\text { type } \mathrm{O} \cap \mathrm{Rh}-)=1-0.06=0.94\)

      94% of people do not have type O, Rh- blood

    105.

    1. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
    2. \(P(C \cup N) = P(C) + P(N) - P(C \cap N) = 0.36 + 0.12 - 0.08 = 0.40\)
    3. \(P(\text { NElTHER chocolate NOR nuts) }=1-P(C \cup N)=1-0.40=0.60\)

    107.

    0

    109.

    \(\frac{10}{67}\)

    111.

    \(\frac{10}{34}\)

    113.

    d

    115.

    1. Race and sex1–1415–2425–64Over 64TOTALS
      White, male2103,36013,6104,87022,050
      White, female805803,3808904,930
      Black, male104601,0601401,670
      Black, female04027020330
      All others 100
      TOTALS3104,65018,7806,02029,760

      Table3.24

    2. Race and sex1–1415–2425–64Over 64TOTALS
      White, male2103,36013,6104,87022,050
      White, female805803,3808904,930
      Black, male104601,0601401,670
      Black, female04027020330
      All others10210460100780
      TOTALS3104,65018,7806,02029,760

      Table3.25

    3. \(\frac{22,050}{29,760}\)
    4. \(\frac{330}{29,760}\)
    5. \(\frac{2,000}{29,760}\)
    6. \(\frac{23,720}{29,760}\)
    7. \(\frac{5,010}{6,020}\)

    117.

    b

    119.

    1. \(\frac{26}{106}\)
    2. \(\frac{33}{106}\)
    3. \(\frac{21}{106}\)
    4. \(\left(\frac{26}{106}\right)+\left(\frac{33}{106}\right)-\left(\frac{21}{106}\right)=\left(\frac{38}{106}\right)\)
    5. \(\frac{21}{33}\)

    121.

    a