In a town, 50% of the families read newspaper X, 40% read newspaper Y…

The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Let X and Y also stand for the events “the family reads X” and “the family reads Y”. Then P(X) = 0.5, P(Y) = 0.4 and \(P(X\cap Y) = 0.2\). By the addition theorem:

    \(P(X\cup Y) = 0.5+0.4-0.2 = 0.7\)

Each item is a conditional probability: the probability of both events divided by the probability of the given event.

(A) Probability that the family reads Y, given that it reads X

This matches (II) \(\frac{2}{5}\).

    \( \displaystyle P(Y|X) = \frac{0.2}{0.5} = \frac{2}{5} \)

(B) Probability that the family reads X, given that it reads Y

This matches (IV) \(\frac{1}{2}\).

    \( \displaystyle P(X|Y) = \frac{0.2}{0.4} = \frac{1}{2} \)

(C) Probability that the family reads X, given that it reads at least one of the two newspapers

This matches (I) \(\frac{5}{7}\).

Every family that reads X reads at least one newspaper, so \(X\cap(X\cup Y) = X\):

    \( \displaystyle P(X|X\cup Y) = \frac{0.5}{0.7} = \frac{5}{7} \)

(D) Probability that the family reads neither newspaper, given that it does not read X

This matches (III) \(\frac{3}{5}\).

A family that reads neither newspaper certainly does not read X, so the probability of both events is the probability of reading neither:

    \(P(X'\cap Y') = 1-0.7 = 0.3\)
    \(P(X') = 1-0.5 = 0.5\)

    \( \displaystyle P(X'\cap Y'\,|\,X') = \frac{0.3}{0.5} = \frac{3}{5} \)