- 11.2 The probability that a single child is albino is 1/4. The probability
that two children are albino is (1/4)(1/4)=1/16. The probability that neither is
albino is (3/4)(3/4)= 9/16.
- 11.4
- a) The probability that none of the five calls reaches a person is
(1-.2)5 = .32768
- b) (1-.08)5= .659
- 11.6
- a) The wheels are independent, so the probability of winning the jackpot
is the product of the probability seeing a bell on each wheel (1/20)(9/20)(1/20).
- b) The probability of seeing two bells on the outer wheels with something
else on the middle wheel is (1/20)(11/20)(1/20). The probability of seeing a bell
on the middle wheel and either of the two outer wheels (but not both of them) is
(19/20)(9/20)(1/20). So, the probability of seeing two bells is
(1/20)(11/20)(1/20)+(19/20)(9/20)(1/20)+(19/20)(9/20)(1/20) = 353/8000.
- 11.7
- b) 30% of college students like country music but not gospel.
- c) 40% of college students like neither country music nor gospel.
- 11.9
- a) The probability of choosing a woman from the population of degree
recipients is 922/1654.
- b) The conditional probability of choosing a woman given that the person
chosen received a professional degree is 32/72.
- 11.12 4% of all adults go to a health club at least twice a week.
P(Belong to health club and Go to club at least twice a week)= P(Belong to club)*
P(Go at least twice a week | Belong to club) = .1 * .4 = .04.
- 11.14
- a) The probability that the first card is a spade is 13/52. The probability
that the second card is a spade given that the first card is a spade is 12/51. The
conditional probabilities for the third, fourth, and fifth cards being spades given
that all previous cards were spades are 11/50, 10/49, and 9/48 respectively.
- b) The general probability rule implies that the probability of getting five
spades in five cards is the product of the conditional probabilities from part a).
The probability is .000495.
- c)The probability of being dealt a flush is 4*.000495=.002.
- 11.17
- b) P(positive test)= (.01)(.9985)+(.99)(.006) = .015925
- 11.18 P(have antibody | positive test) = (.01)(.9985)/.015925 = .627
- 11.20 P(win at least once) = 1- P(never winning) = 1 - .985 = .096
- 11.23
- a) The probability of choosing a man from the population of degree recipients is
732/1654.
- b)P(Bachelor's degree | Male recipient) = 505/732
- c) P(Male and Bachelor's Degree) = (732/1654)(505/732)= 505/1654.
- 11.29
- a) P(drawing a bad switch) = 1/10. P(drawing a good switch)=9/10.
- b) If one switch is drawn and it is bad, there are 9999 switches remaining with
999 of them bad. P(second switch drawn bad | first switch bad) = 999/9999.
- c) If one switch is drawn and it is good, there are 9999 switches remaining with
1000 of them bad. P(second switch drawn bad | first switch good) = 1000/9999.
- 11.41 The probability that Maria's husband has type B or type O blood is .45+.11=.56
- 11.42 That a husband and wife share the same blood type is
.452+.42+.112+.042= .3762
- 11.43
- a) P(wife type A and husband type B)= .4*.11= .044.
- b) The probability that one of the couple has type A blood and the other has type
B is .088.
- 11.44
| Blood Type |
|---|
| O | A | B | AB |
| Rh | + | .378 | .336 | .0924 | .0336 |
|---|
| - | .072 | .064 | .0176 | .0064 |
- 11.45 P(at least one of 10 people is a O-negative)=1-P(none are O-negative)=
1-(.928)10= .5263.
- 11.48
- a) P(first is type A and second is type A)= (.5)(.5)=.25.
- b) P(both have same blood type)=P(both type A or both type B or both type AB or
both type O)= (.5)(.5)+(.25)(.25)+(.25)(.25)+ 0 = .375.