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 number is greater than 11	(i) 0
(B) The sum of the number is 4 or less	(ii) 1/15
(C) The sum of the number is 4	(iii) 2/15
(D) The sum of the number is 4	(iv) 3/15

Choose the correct answer from the option given below

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

Answer 1

Answer: (A)

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

Answer 2

(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. Hence, $(A) \to (I)$.

(B) The sum of the numbers is 4 or less: The possible pairs are $(1,2), (2,1), (1,3), (3,1), (2,2)$, but $(2,2)$ is excluded, leaving 4 favorable outcomes. Out of the 30 possible outcomes (only different numbers), the probability is:

$P = \frac{4}{30} = \frac{1}{15}.$

Hence, $(B) \to (II)$.

(C) 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:

$P = \frac{2}{30} = \frac{2}{15}.$

Hence, $(C) \to (III)$.

(D) The sum of the numbers is 7: The possible pairs are $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$, all of which involve different numbers, giving 6 favorable outcomes. Thus, the probability is:

$P = \frac{6}{30} = \frac{3}{15}.$

Hence, $(D) \to (IV)$.