Let A and B be independent events such that P(A) = 3/7 and P(B) = 1/7, Then Match List-I with List-II, List-I, List-II, (A) P(A and B), (I) 24/49, (B) P(A or B), (II) 18/49, (C) P(A and not B), (III) 25/49, (D) P(neither A nor B), (IV) 3/49, Choose the correct answer from the options given below: Option 1: (A) - (IV), (B) - (III), (C) - (II), (D) - (I) Option 2: (A) - (IV), (B) - (III), (C) - (I), (D) - (II) Option 3: (A) - (II), (B) - (III), (C) - (IV), (D) - (I) Option 4: (A) - (II), (B) - (III), (C) - (I), (D) - (IV)

Answer: (A) - (IV), (B) - (III), (C) - (II), (D) - (I). Solution: Independence gives P(A B)=3717=3/49 (IV); P(A B)=37+17-3/49=25/49 (III); P(A B')=3767=18/49 (II); P(A' B')=47*67=24/49 (I).

Algebra

Probability

Easy

core

Let A and B be independent events such that $P(A) = \frac{3}{7}$ and $P(B) = \frac{1}{7}$ , Then

Match List-I with List-II

List-I List-II
(A) P(A and B) (I) $\frac{24}{49}$
(B) P(A or B) (II) $\frac{18}{49}$
(C) P(A and not B) (III) $\frac{25}{49}$
(D) P(neither A nor B) (IV) $\frac{3}{49}$

Choose the correct answer from the options given below:

List-I	List-II
(A) P(A and B)	(I) $\frac{24}{49}$
(B) P(A or B)	(II) $\frac{18}{49}$
(C) P(A and not B)	(III) $\frac{25}{49}$
(D) P(neither A nor B)	(IV) $\frac{3}{49}$

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

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

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

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

✅ Correct Option: 1

Independence gives $P(A\cap B)=\frac37\cdot\frac17=\frac{3}{49}$ (IV); $P(A\cup B)=\frac37+\frac17-\frac{3}{49}=\frac{25}{49}$ (III); $P(A\cap B')=\frac37\cdot\frac67=\frac{18}{49}$ (II); $P(A'\cap B')=\frac47\cdot\frac67=\frac{24}{49}$ (I).

2026: 15 May Shift 1

Let A and B are two events associated to a random experiment, then

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $P(A \cap B) + P(A \cap \overline{B})$	I. $P(A \cup B)$
B. $P(A \cap B) + P(\overline{A} \cap B)$	II. $P(A) + P(B)$
C. $P(A \cap B) + P(\overline{A} \cap B) + P(A \cap \overline{B})$	III. $P(A)$
D. $P(A \cup B) + P(A \cap B)$	IV. $P(B)$

Choose the correct answer from the options given below:

2026: 12 May Shift 2

Given that the events A and B are mutually exclusive such that $P(A)=\dfrac{1}{2}$ , $P(A\cup B)=\dfrac{2}{3}$ and $P(B)=p$ , then the value of $p$ is:

2025: 22 May Shift 1

Two persons A and B throw a die alternately till one of them gets a six and wins the game. If A begins, then the probabilities of winning of A and B respectively are

2026: 21 May Shift 1

A fair coin is tossed until a head or five tails occur. The probability distribution of the number of tosses is,

2025: 22 May Shift 1

Let X denotes the number of doublets obtained in 3 throws of a pair of dice.

Match List-I with List-II

List-I	List-II
(A) $P(X = 0)$	(I) $\frac{1}{216}$
(B) $P(X = 1)$	(II) $\frac{15}{216}$
(C) $P(X = 2)$	(III) $\frac{75}{216}$
(D) $P(X = 3)$	(IV) $\frac{125}{216}$

Choose the correct answer from the options given below:

2026: 18 May Shift 2

If A and B are independent events such that $P(A)=\dfrac{1}{2}$ , $P(A\cup B)=\dfrac{3}{5}$ and $P(B)=p$ , then value of p is:

Solution

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