Let A and B are two events associated to a random experiment, then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( P(A\cap B)+P(A\cap\bar{B}) \) | I. | \( P(A\cup B) \) |
| B. | \( P(A\cap B)+P(\bar{A}\cap B) \) | II. | \( P(A)+P(B) \) |
| C. | \( P(A\cap B)+P(\bar{A}\cap B)+P(A\cap\bar{B}) \) | III. | \( P(A) \) |
| D. | \( P(A\cup B)+P(A\cap B) \) | IV. | \( P(B) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-IV, C-I, D-II
A splits into the disjoint parts \( A\cap B \) and \( A\cap\bar{B} \); similarly for B.
(A) \( P(A\cap B)+P(A\cap\bar{B}) \)
This matches III. P(A). The two parts together make up A.
(B) \( P(A\cap B)+P(\bar{A}\cap B) \)
This matches IV. P(B). The two parts together make up B.
(C) \( P(A\cap B)+P(\bar{A}\cap B)+P(A\cap\bar{B}) \)
This matches I. P(A ∪ B). The three disjoint parts (both, only B, only A) together make up \( A\cup B \).
(D) \( P(A\cup B)+P(A\cap B) \)
This matches II. P(A) + P(B). By the addition theorem,
\( \displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
so adding \( P(A\cap B) \) gives \( P(A)+P(B) \).