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Stats 13.1 HW 2 Solutions
October 10, 2007
a) The probability that both are affected is 0% because female offspring have no chance of getting the disease.
b) The probability that only one sibling is affected can be written as: P(one child affected) = P(male affected and female not) + P(female affected and male not) = 50% · 100% + 0% · 50% = 50%
a)
P(get question right) = P(studied the question) + P(didn’t study the question & got it right) = 40% + 60% · 20% = 52%
b)
P(studied it — got it right) = P(studied question & got it right) P(got it right) = 40% 52% = 76 .9%
a) There are two ways to test positive, a true positive and a false positive. P(test positive) = P(true positive) + P(false positive) = 0.
b)
P(has disease — test positive) = P(has disease & test positive) P(test positive) = P(true positive) P(test positive) = 00 ..^092146 = 0. 63
The law of independence states:
A and B are independent if and only if P(A+B) = P(A)·P(B).
But, P(husband and wife smoke) =? P(husband smokes)·P(wife smokes) 8% =? 30% · 20% 8% 6 = 6% So the smoking status of the husband is not independet of the status of the wife.
5 3.18 and 3.
a)
P(Y = 3) = # of broods with Y = 3 total # of broods = 5000610 = 0. 122
a) Let us consider a success as “high blood lead level.” Then we have a binomial problem with n = 16 and p = 1/8. Then, P(none have high blood lead level) = P(no successes) = 16 C 0 ·
8
b) P(one has high blood lead level) = P(1 success) = 16 C 1 ·
8
8
c) P(two have high blood lead level) = P(2 successes) = 16 C 2 ·
8
8
d) P(three or more have high blood lead level) = 1 - P(2 or fewer have high blood lead level) = 1 - (P(0 have high blood lead level) + P(1 have high blood lead level) + P(2 have high blood lead level)) = 1 - (0.1181 + 0.2699 + 0.2891) = 0.
