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

To solve the problem of matching List-I with List-II based on the probability of the events described, we consider the conditions for rolling a pair of dice:

1. Total possible outcomes when rolling two dice = 6 x 6 = 36.

2. We are interested in scenarios where the numbers on both dice are different.

3. Outcomes where numbers on both dice are the same (i.e., doubles) = 6 (e.g., (1,1), (2,2), ..., (6,6)).

4. Outcomes with different numbers = 36 - 6 = 30.

We analyze the given events:

LIST-I(EVENT)	Valid Outcomes	LIST-II(PROBABILITY)
(A) The sum of the number is greater than 11	(5,6), (6,5)	0
(B) The sum of the number is 4 or less	(1,2), (2,1), (1,3), (3,1)	1/15
(C) The sum of the number is 4	(1,3), (3,1), (2,2)	2/15
(D) The sum of the number is 4	(1,3), (3,1), (2,2)	3/15

5. Each probability is computed using the formula Number of favorable outcomes / Number of different outcomes.

6. Calculating probabilities for valid outcomes:

- Event (A) probability = 0/30 = 0 (because outcomes are invalid since both numbers appear are different).

- Event (B) probability = 2/30 = 1/15.

- Event (C) probability = 4/30 = 2/15.

- Event (D) probability = 4/30 = 3/15.

Thus, the correct matching is: (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, \( (D) \to (IV) \).