Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

A pair of dice is rolled. If the two numbers appearing on them are different, the probability that Match List-I with List-II.

List-I (Event)

List-II (Probability)

(A) The sum of the numbers is greater than 11

(I) 0

(B) The sum of the numbers is 4 or less

(II) 1/15

(C) The sum of the numbers is 4

(III) 2/15

(D) The sum of the numbers is 7

(IV) 3/15

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

List-I (Event)

List-II (Probability)

(A) The sum of the numbers is greater than 11

(I) 0

(B) The sum of the numbers is 4 or less

(III) 2/15

(C) The sum of the numbers is 4

(II) 1/15

(D) The sum of the numbers is 7

(IV) 3/15

When two dice are rolled:
Total possible outcomes = 6 × 6 = 36
But pairs like (1,1), (2,2), ..., (6,6) (i.e., 6 repeated pairs) are not allowed.

So, valid outcomes = 36 - 6 = 30

(A) The sum of the numbers is greater than 11: The only possible pair for a sum greater than 11 is (6,6), which is excluded as the numbers must be different. Thus, the probability is 0. 

$P(A)=0$,

(B) Sum of numbers 4 or less are,

$1+2=3$

$2+1=3$

$1+3=4$

$3+1=4$

Probability = $\frac{4}{30}=\frac{2}{15}$

(C) Sum of number in 4 : The sum of the numbers is 4: The possible pairs are (1,3),(3,1),(2,2), but (2,2) is excluded, leaving 2 favorable outcomes. Thus, the probability is:

Probability = $\frac{2}{30}=\frac{1}{15}$

(D) Sum of the number is 7

$1+6,6+1$

$2+5,5+2$

$3+4,4+3$

Probability = $\frac{6}{30}=\frac{3}{15}$