Question

Match List I with List II.
Let A and B be events with P(A)=⅔, P(B)=½ and p(A∩B)=⅓.

List I Probability of an event		List II  Value
(A)	P(A∩Bc)	(I)	\(\frac{2}{3}\)
(B)	P(A∪Bc)	(II)	\(\frac{1}{3}\)
(C)	P(Ac∩Bc)	(III)	\(\frac{5}{6}\)
(D)	P(Ac∪Bc)	(IV)	\(\frac{1}{6}\)

(Here c stands for complement)
Choose the correct answer from the options given below:

• A. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• B. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
• C. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
• D. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Answer 1

The Correct Option is A

We are given two events \( A \) and \( B \) with probabilities \( P(A) = \frac{2}{3} \), \( P(B) = \frac{1}{2} \), and \( P(A \cap B) = \frac{1}{3} \). We need to find the probabilities for several combinations of these events and match them to the given values.

1. Find \( P(A \cap B^c) \):

   We have the formula \( P(A \cap B^c) = P(A) - P(A \cap B) \).

   Substituting the known values: \[ P(A \cap B^c) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \] Hence, (A) matches with (II).

2. Find \( P(A \cup B^c) \):

   The formula for computing \( P(A \cup B^c) \) is: \[ P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) \] Where \( P(B^c) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2} \).

   Calculating \( P(A \cup B^c) \): \[ P(A \cup B^c) = \frac{2}{3} + \frac{1}{2} - \frac{1}{3} \] Converting to a common denominator (6): \[ P(A \cup B^c) = \frac{4}{6} + \frac{3}{6} - \frac{2}{6} = \frac{5}{6} \] Hence, (B) matches with (III).

3. Find \( P(A^c \cap B^c) \):

   Using the formula: \[ P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B)) \] Substitute the given values: \[ P(A^c \cap B^c) = 1 - \left(\frac{2}{3} + \frac{1}{2} - \frac{1}{3}\right) \] Calculating: \[ P(A^c \cap B^c) = 1 - \left(\frac{3 + 4 - 2}{6}\right) = 1 - \frac{5}{6} = \frac{1}{6} \] Hence, (C) matches with (IV).

4. Find \( P(A^c \cup B^c) \):

   We have already found that: \[ P(A^c \cup B^c) = P(A \cap B^c) + P(B \cap A^c) + P(A^c \cap B^c) = P(A \cup B)^c \] From \( P(A) + P(B) - P(A \cap B) \), it's \(\frac{5}{6}\). so: \[ P(A^c \cup B^c) = P(A \cap B)^c = \frac{2}{3} \] Hence, (D) matches with (I).

Based on calculations, the correct matching is:

• (A) - (II)
• (B) - (III)
• (C) - (IV)
• (D) - (I)