Question

Let A and B be two events such that: \[ P(A) = 0.8, \quad P(B) = 0.5, \quad P(B|A) = 0.4 \]

Match List-I with List-II:

List-I	List-II
(A) \(P(A \cap B)\)	(I) 0.2
(B) \(P(A|B)\)	(II) 0.32
(C) \(P(A \cup B)\)	(III) 0.64
(D) \(P(A')\)	(IV) 0.98

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

Hint:

The first step in many conditional probability problems is to find $P(A \cap B)$. It can usually be found from one of the two multiplication rules: $P(A \cap B) = P(A|B)P(B) = P(B|A)P(A)$. Once you have the intersection probability, most other quantities can be calculated.

Answer 1

The Correct Option is B

Step 1: Conceptual Understanding:
This problem requires the application of fundamental probability rules: the multiplication rule for conditional probability, the definition of conditional probability, the addition rule, and the complement rule.
Step 3: Detailed Calculation:
Given: P(A) = 0.8, P(B) = 0.5, P(B|A) = 0.4
(A) Probability of A and B: P(A $\cap$ B)
Applying the multiplication rule: $P(A \cap B) = P(B|A) \times P(A)$.
$P(A \cap B) = 0.4 \times 0.8 = 0.32$. This result aligns with option (II).
(B) Conditional Probability of A given B: P(A|B)
Using the formula for conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
$P(A|B) = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{25} = 0.64$. This result aligns with option (III).
(C) Probability of A or B: P(A $\cup$ B)
Applying the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.8 + 0.5 - 0.32 = 1.3 - 0.32 = 0.98$. This result aligns with option (IV).
(D) Probability of not A: P(A')
Using the complement rule: $P(A') = 1 - P(A)$.
$P(A') = 1 - 0.8 = 0.2$. This result aligns with option (I).
Step 4: Conclusion:
The correct pairings are:
(A) $\rightarrow$ (II)
(B) $\rightarrow$ (III)
(C) $\rightarrow$ (IV)
(D) $\rightarrow$ (I)
This corresponds to choice (2).