Match List-I with List-II
|
List-I (Statements/Expressions, etc.) |
List-II (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 $E_2$, $P(E_1 ∪ E_2)=$ |
(III) $1-P(\vec E)$ |
|
(D) For independent events A and B, $P(A ∩ B)=$ |
(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)
|
List-I (Statements/Expressions, etc.) |
List-II (Value/Expressions, etc.) |
|
(A) $P(E)$ |
(III) $1-P(\vec E)$ |
|
(B) The probability of an impossible event |
(IV) 0 |
|
(C) For exhaustive events $E_1$ and $E_2$, $P(E_1 ∪ E_2)=$ |
(I) 1 |
|
(D) For independent events A and B, $P(A ∩ B)=$ |
(II) $P(A).P(B)$ |
(A) $P(E)$ — (III) $1 - P(\bar{E})$
According to the rule of complementary events, the probability of an event $E$ occurring plus the probability of it not occurring ($\bar{E}$) is always equal to $1$. Therefore, $P(E) = 1 - P(\bar{E})$.
(B) The probability of an impossible event — (IV) 0
An impossible event is one that cannot happen under any circumstances. By definition, its probability is always $0$.
(C) For exhaustive events $E_1$ and $E_2$, $P(E_1 \cup E_2) =$ — (I) 1
Exhaustive events are those that cover all possible outcomes of a sample space. When $E_1$ and $E_2$ are exhaustive, their union ($E_1 \cup E_2$) represents the entire sample space, so the probability of either occurring is $1$.
(D) For independent events $A$ and $B$, $P(A \cap B) =$ — (II) $P(A) \cdot P(B)$
The definition of independent events is that the occurrence of one does not affect the other. Mathematically, the probability of both occurring simultaneously ($A \cap B$) is the product of their individual probabilities.