Let A and B be two events such that \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.1\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(P(A \cup B)\) | (I) | 0.2 |
| (B) | \(P(A|B)\) | (II) | 0.8 |
| (C) | \(P(B|A)\) | (III) | 0.4 |
| (D) | \(P(A \cap B')\) | (IV) | 0.25 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Here \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.1\).
(A) \(P(A \cup B)\)
This matches (II) 0.8.
By the addition rule:
\( P(A \cup B) = 0.5 + 0.4 - 0.1 = 0.8 \)
(B) \(P(A|B)\)
This matches (IV) 0.25.
The given event B goes in the denominator:
\( \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.4} = 0.25 \)
(C) \(P(B|A)\)
This matches (I) 0.2.
Now the given event is A:
\( \displaystyle P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.5} = 0.2 \)
(D) \(P(A \cap B')\)
This matches (III) 0.4.
This is the part of A outside B:
\( P(A \cap B') = P(A) - P(A \cap B) = 0.5 - 0.1 = 0.4 \)