In a study of alcoholics, it was found that 40% had a alcoholic fathers and 6% h
ID: 2929850 • Letter: I
Question
In a study of alcoholics, it was found that 40% had a alcoholic fathers and 6% had alcoholic mothers. Forty-two percent percent had a least one alcoholic parent.
1. What is the probability that a randomly selected alcoholic will have both parents alcoholic?
2. What is the probability that a randomly selected alcoholic will have an alcoholic mother if the father is alcoholic?
3. What is the probability that a randomly selected alcoholic will have an alcoholic mother but not an alcoholic father?
4. What is the probability that a randomly selected alcoholic will have an alcoholic mother if the father is not alcoholic?
Explanation / Answer
Let A denote the event of having an alcoholic father and B denote the event of having an alcoholic mother.
P(A) = 0.4 P(B) = 0.06
P(A U B) = 0.42
1. The probability that a randomly selected alcoholic will have both parents alcoholic
= P(A B) = P(A) + P(B) - P(A U B)
= 0.4 + 0.06 - 0.42
= 0.04.
2. The probability that a randomly selected alcoholic will have an alcoholic mother if the father is alcoholic
P(B|A) = P(A B) / P(A)
= 0.04 / 0.4
= 0.1.
3. The probability that a randomly selected alcoholic will have an alcoholic mother but not an alcoholic father
P(A' B) = P(B) - P(A B)
= 0.06 - 0.04
= 0.02.
4. The probability that a randomly selected alcoholic will have an alcoholic mother if the father is not alcoholic
= P(B' | A') = P(A' B') / P(A') = P(A U B)' / (1 - P(A)) = (1 - P(A U B)) / (1 - P(A))
= (1 - 0.42) / (1 - 0.4)
= 0.58 / 0.6
= 0.9667.