Match List - I with List - II.
| List - I | List - II | ||
| (A) | \(P(\overline{A} \cap B)\) | (I) | \(P(A) + P(B)\) |
| (B) | \(P(A \cap \overline{B})\) | (II) | \(P(A) + P(B) - 2P(A \cap B)\) |
| (C) | \(P\left[(A \cap \overline{B}) \cup (\overline{A} \cap B)\right]\) | (III) | \(P(B) - P(A \cap B)\) |
| (D) | \(P(A \cup B) + P(A \cap B)\) | (IV) | \(P(A) - P(A \cap B)\) |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
B is made of two disjoint parts: the part inside A and the part outside A. The same holds for A:
\(P(B) = P(A \cap B) + P(\overline{A} \cap B)\)
\(P(A) = P(A \cap B) + P(A \cap \overline{B})\)
(A) \(P(\overline{A} \cap B)\)
This matches (III) \(P(B) - P(A \cap B)\).
\(P(\overline{A} \cap B) = P(B) - P(A \cap B)\)
(B) \(P(A \cap \overline{B})\)
This matches (IV) \(P(A) - P(A \cap B)\).
\(P(A \cap \overline{B}) = P(A) - P(A \cap B)\)
(C) \(P\left[(A \cap \overline{B}) \cup (\overline{A} \cap B)\right]\)
This matches (II) \(P(A) + P(B) - 2P(A \cap B)\).
The two events are disjoint (one lies inside B, the other outside B), so their probabilities add:
\(= \left[P(A) - P(A \cap B)\right] + \left[P(B) - P(A \cap B)\right]\)
\(= P(A) + P(B) - 2P(A \cap B)\)
(D) \(P(A \cup B) + P(A \cap B)\)
This matches (I) \(P(A) + P(B)\).
By the addition theorem, \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), so
\(P(A \cup B) + P(A \cap B) = P(A) + P(B)\)