Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Statements/Expressions, etc.) | (Value/Expressions, etc.) | ||
| (A) | \(P(E)\) | (I) | 1 |
| (B) | The probability of an impossible event | (II) | \(P(A).P(B)\) |
| (C) | For exhaustive events \(E_1\) and \(E_2\), \(P(E_1\cup E_2)=\) | (III) | \(1-P(\bar{E})\) |
| (D) | For independent events A and B, \(P(A\cap B)=\) | (IV) | 0 |
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)-(I), (D)-(II)
Each List-I item is a basic result of probability. Here \(\bar{E}\) is the event ‘not E’, the complement of E.
(A) \(P(E)\) — this matches (III) \(1-P(\bar{E})\). An event either happens or does not, so \(P(E)+P(\bar{E})=1\), which gives \(P(E)=1-P(\bar{E})\).
(B) The probability of an impossible event — this matches (IV) 0. An impossible event has no favourable outcome, so its probability is \(\frac{0}{n}=0\).
(C) For exhaustive events \(E_1\) and \(E_2\), \(P(E_1\cup E_2)=\) — this matches (I) 1. Exhaustive events between them cover every possible outcome, so one of them is sure to happen and their union is the sure event.
(D) For independent events A and B, \(P(A\cap B)=\) — this matches (II) \(P(A).P(B)\). Events are independent when one happening does not change the chance of the other, and then the probability that both happen is the product of their probabilities.