Let A and B are two events such that P(A) = 0.8, P(B) = 0.5, P(B—A) = 0.4 Match List-I with List-II

List-I	List-II
(A) $P(A\cap B)$ (B) $P(A-B)$ (C) $P(A\cup B)$ (D) $P\left(A^{\text{'}}\right)$	(I) 0.2 (II) 0.32 (III) 0.64 (IV) 0.98

Choose the correct answer from the options given below:

The correct option is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).

Step-by-Step Explanation:
We are given the following probabilities:
P(A) = 0.8
P(B) = 0.5
The term P(B—A) represents the conditional probability of event B given event A, commonly written as P(B|A) or P(B/A). Thus, we have:
P(B|A) = 0.4

Now, we calculate the matching values for each item in List-I:

1. Finding (A) P(A ∩ B):
By the multiplication rule of probability, we know that:
$$P(A\cap B)=P(A)\cdot P(B|A)$$
Substituting the given values:
$$P(A\cap B)=0.8\cdot 0.4=0.32$$
Thus, (A) matches with (II).

2. Finding (B) P(A|B) (notated in List-I as P(A-B) or P(A/B)):
Using the definition of conditional probability:
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
Substituting our calculated value of P(A ∩ B) and the given value of P(B):
$$P(A|B)=\frac{0.32}{0.5}=0.64$$
Thus, (B) matches with (III).

3. Finding (C) P(A ∪ B):
Using the addition theorem of probability:
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
Substituting the values:
$$P(A\cup B)=0.8+0.5-0.32=1.3-0.32=0.98$$
Thus, (C) matches with (IV).

4. Finding (D) P(A'):
Using the complement rule:
$$P(A^{\prime })=1-P(A)$$
Substituting the given value of P(A):
$$P(A^{\prime })=1-0.8=0.2$$
Thus, (D) matches with (I).