Question

The probability that a married man watches,a certain t.v show is 0.4 and the

probability that a married woman watches the show is 0.5,the probability that a
man watches the show, given that his wife does is 0.7 find
(A) the probability that a married couple watches the show.
(B) the probability that a wife watches the show given that her husband does.
(C) the probability that at least 1 person of married couple will watch the
show.

Answer 1

The probability that both a married man and woman will watch the show is 20%. The probability that a wife watches the show given that her husband does is 87.5%. The probability that at least one person in a married couple will watch the show is 70%.

Step-by-step explanation:

To solve the given probability problems, we'll denote the probability of a man watching the show as P(M), a woman as P(W), a man watching given that his wife does as P(M|W), and the couple watching the show as P(M&W). The following solutions are based on the principles of conditional probability and independence.

(A) Probability that a married couple watches the show

The probability that both a married man and woman will watch the show, assuming independence, would be the product of their individual probabilities:

P(M&W) = P(M) * P(W) = 0.4 * 0.5 = 0.20 or 20%

(B) Probability that a wife watches the show given that her husband does

Using Bayes' theorem:

P(W|M) = P(M|W) * P(W) / P(M) = 0.7 * 0.5 / 0.4 = 0.875 or 87.5%

(C) Probability that at least one person of a married couple will watch the show

To find the probability that at least one person watches the show, we calculate 1 minus the probability that neither watches. As these events are independent:

P(neither watches) = (1 - P(M)) * (1 - P(W)) = (1 - 0.4) * (1 - 0.5) = 0.6 * 0.5 = 0.3

So, P(at least one watches) = 1 - P(neither watches) = 1 - 0.3 = 0.70 or 70%