Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Probability

Question:

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:

Options:

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

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.