Algebra

Probability

Medium

common

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:

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) $\frac{125}{216}$

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

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

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

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

✅ Correct Option: 3

A doublet occurs when both dice show the same number (1-1, 2-2, 3-3, 4-4, 5-5, or 6-6). The variable X represents the number of doublets obtained in 3 throws of a pair of dice.

When throwing a pair of dice once:

Total outcomes = 6 × 6 = 36

Doublet outcomes = 6

Non-doublet outcomes = 30

Therefore:

P(doublet) = $\frac{6}{36}$

$= \frac{1}{6}$

P(not a doublet) = $\frac{30}{36}$

$= \frac{5}{6}$

Since the throw is repeated 3 times independently, this follows a binomial distribution with:

n = 3 (number of trials)

p = $\frac{1}{6}$ (probability of doublet)

q = $\frac{5}{6}$ (probability of no doublet)

The probability formula is:

$P(X = k) = C(3,k) \times \left(\frac{1}{6}\right)^k \times \left(\frac{5}{6}\right)^{3-k}$

For P(X = 0):

$P(X = 0) = C(3,0) \times \left(\frac{1}{6}\right)^0 \times \left(\frac{5}{6}\right)^3$

$= 1 \times 1 \times \frac{125}{216}$

$= \frac{125}{216}$ → (IV)

For P(X = 1):

$P(X = 1) = C(3,1) \times \left(\frac{1}{6}\right)^1 \times \left(\frac{5}{6}\right)^2$

$= 3 \times \frac{1}{6} \times \frac{25}{36}$

$= \frac{75}{216}$ → (III)

For P(X = 2):

$P(X = 2) = C(3,2) \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^1$

$= 3 \times \frac{1}{36} \times \frac{5}{6}$

$= \frac{15}{216}$ → (II)

For P(X = 3):

$P(X = 3) = C(3,3) \times \left(\frac{1}{6}\right)^3 \times \left(\frac{5}{6}\right)^0$

$= 1 \times \frac{1}{216} \times 1$

$= \frac{1}{216}$ → (I)

The correct matching is:

(A) P(X = 0) = $\frac{125}{216}$ → (IV)

(B) P(X = 1) = $\frac{75}{216}$ → (III)

(D) P(X = 3) = $\frac{1}{216}$ → (I)

2026: 21 May Shift 1

A problem is given to 3 students A,B and C, whose probabilities of solving are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively, then

(A). P(Problem solved) = $\frac{3}{4}$

(B). P(Problem not solved) = $\frac{1}{4}$

(C). P(Problem solved by exactly one student) = $\frac{7}{24}$

(D). P(Problem solved by at least 2 students) = $\frac{17}{24}$

Choose the correct answer from the options given below:

2022: 17 Aug Shift 2

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2023: 23 May Shift 3

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. Then variance of the number of kings is

2026: 18 May Shift 2

An alphabet has been selected from English alphabets. What is the probability that it is a vowel given that it precedes 'M' ?

2025: 15 May Shift 1

The probability of a man hitting a target is 1/2. How many times must he fire so that the probability of hitting the target at least once is more than 90%?

2025: 14 May Shift 1

If A is any event associated with sample space and If $E_1, E_2, E_3$ are mutually exclusive and exhaustive events. Then which of the following are true?

(A) $P(A) = P(E_1)P(E_2|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$

(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$

(D) $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{i=1}^3 P(E_i|A)P(E_i)}$ , $i = 1,2,3$

Choose the correct answer from the options given below:

Solution

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