Question

Match List I with List II. Let A and B be events with $P(A) = \frac{2}{3}, P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{3}$

List I: (A) $P(A \cap B^c)$ (B) $P(A \cup B^c)$ (C) $P(A^c \cap B^c)$ (D) $P(A^c \cup B^c)$

List II: (I) $\frac{2}{3}$ (II) $\frac{1}{3}$ (III) $\frac{5}{6}$ (IV) $\frac{1}{6}$

Choose the correct answer from the options given below:

Accepted Answer

A. - (A) $P(A \cap B^c) = P(A) - P(A \cap B) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \implies$ (II) - (B) $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = \frac{2}{3} + \frac{1}{2} - \frac{1}{3} = \frac{5}{6} \implies$ (III) - (C) $P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - \left(\frac{2}{3} + \frac{1}{2} - \frac{1}{3}\right) = 1 - \frac{5}{6} = \frac{1}{6} \implies$ (IV) - (D) $P(A^c \cup B^c) = 1 - P(A \cap B) = 1 - \frac{1}{3} = \frac{2}{3} \implies$ (I) Correct match: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).

Match probability lists

Explanation

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