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

The Correct Option is A

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