Algebra

Probability

Easy

core

Match List-I with List-II

if $P(A) = \frac{3}{7}$ , $P(B) = \frac{4}{7}$ and $P(A \cup B) = \frac{5}{7}$

List-I List-II
(A) $P(A \cap B)$ (I) $\dfrac{2}{3}$
(B) $P(A \mid B)$ (II) $\dfrac{5}{7}$
(C) $P(B \mid A)$ (III) $\dfrac{1}{2}$
(D) $P(A' \cup B')$ (IV) $\dfrac{2}{7}$

Choose the correct answer from the options given below:

List-I	List-II
(A) $P(A \cap B)$	(I) $\dfrac{2}{3}$
(B) $P(A \mid B)$	(II) $\dfrac{5}{7}$
(C) $P(B \mid A)$	(III) $\dfrac{1}{2}$
(D) $P(A' \cup B')$	(IV) $\dfrac{2}{7}$

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

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

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

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

✅ Correct Option: 3

Given $P(A) = \frac{3}{7}$ , $P(B) = \frac{4}{7}$ , and $P(A \cup B) = \frac{5}{7}$

Using the Addition Rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\frac{5}{7} = \frac{3}{7} + \frac{4}{7} - P(A \cap B)$

$\frac{5}{7} = \frac{7}{7} - P(A \cap B)$

$P(A \cap B) = \frac{2}{7} \longrightarrow \textbf{(IV)}$

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$

$= \dfrac{\frac{2}{7}}{\frac{4}{7}}$

$= \frac{2}{4}$

$= \frac{1}{2} \longrightarrow \textbf{(III)}$

$P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$

$= \dfrac{\frac{2}{7}}{\frac{3}{7}}$

$= \frac{2}{3} \longrightarrow \textbf{(I)}$

Using De Morgan's Law: $A' \cup B' = (A \cap B)'$

$P(A' \cup B') = 1 - P(A \cap B)$

$= 1 - \frac{2}{7}$

$= \frac{5}{7} \longrightarrow \textbf{(II)}$

List-I	Answer	List-II
(A) $P(A \cap B) = \frac{2}{7}$	→	(IV)
(B) $P(A \mid B) = \frac{1}{2}$	→	(III)
(C) $P(B \mid A) = \frac{2}{3}$	→	(I)
(D) $P(A' \cup B') = \frac{5}{7}$	→	(II)

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

2025: 27 May Shift 1

It is given that 3% of items manufactured by an industry are defective. The probability that a packet of 250 items contains one defective item is: [Given: $e^{-7.5} \approx 0.000553$ ]

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:

2025: 3 June Shift 1

Which of the following statements are correct?

(A) The mean and variance of the Poisson distribution are equal.

(B) The mean and variance of a Binomial distribution are equal.

(C) An unbiased die is thrown again and again until two sixes are obtained, then the probability of obtaining the second six in the 3rd throw is $\frac{5}{108}$ .

(D) If the variance of a Poisson distribution is 2, then P(X = 2) = $2e^{-2}$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

Five bad apples are mixed with 15 good apples accidently. Let the random variable X be the number of bad apples in a draw of two apples, then mean of X is

2025: 21 May Shift 2

If $X$ is a random variable which can assume values $0, 1, 2, 3$ or $4$ such that $P(X = 1) = P(X = 2)$ and $3P(X = 3) = 4P(X = 4) = P(X = 0) = \frac{1}{8}$ , then $P(X > 0)$ is:

2025: 16 May Shift 1

In a game, a man wins ₹ 8 for getting a number greater than 3 and loses ₹ 3 otherwise, when a fair die is thrown. The man decided to throw a die 4 times but to quit as and when he gets a number greater than 3. If X denotes the amount which the man wins or loses, then which of the following are correct?

(A) All the possible values of X are 8, 5, 2 and -1.

(B) The probability distribution of X is:

X	8	5	2	-1	-12
P(X)	1/2	1/4	1/8	1/16	1/16

(D) The variance of X is 6615/256.

Choose the correct answer from the options given below:

Solution

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