(1) A batch of 300 samples of containers for frozen apple juice contains eight t
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Question
(1) A batch of 300 samples of containers for frozen apple juice contains eight that are defective. Three are selected at random, without replacement from the batch (a) What is the probability that the second one selected is defective given that the first one was defective? (b) What is the probability that both are defective? c) What is the probability that both are acceptable? (2) In a presidential election, exit polls from Washington State provided the following results Not employee (60%) Employee (40%) John 42% 51% Bush 58% 49% (a) What is the total probability of voting for Bush? (b) If a randomly selected respondent has voted for John, what is the probability that the person is an employee?Explanation / Answer
(1) (a) Probability that a container is defective = 8/300.
Probability that the second is defective given the first one was defective = 7/299.
(b) Probability that both are defective = = 8/300 * 7/299 = 0.0006.
(c) Probability that both are acceptable = (1 - 8/300) * (1 - 8/299) = 0.9473.
(2) Let A denote voting for Bush and B denote person is an employee.
(a) Probability of voting for Bush if the person is an employee P(A|B)= 0.49
Probability of voting for Bush if the person is not an employee P(A|B')= 0.58
=> P(A B) = P(A|B) P(B) = 0.49*0.4
P(A B') = P(A|B') P(B') = 0.58*0.6
Total probability of voting for Bush P(A) = 0.58*0.60 + 0.49*0.40 = 0.544
(b) Total probability of voting for John P(A') = 1 - 0.544 = 0.456.
P(A' B) = P(B) - P(A B) = 0.4 - 0.49*0.4 = 0.51*0.4
Probability that the person is an employee provided that the person voted for John P(B|A')
= P(A' B) / P(A')
= 0.51*0.4 / 0.456
= 0.4474.
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