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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
The two numbers are given to be different, so the sample space is reduced: of the 36 outcomes, the 6 doubles (1, 1), (2, 2), …, (6, 6) are removed.
\(n(S) = 36 - 6 = 30\)
(A) The sum of the numbers is greater than 11
This matches (I) 0.
The only sum greater than 11 is 12 = 6 + 6, a double, which is excluded:
\( \displaystyle P = \frac{0}{30} = 0 \)
(B) The sum of the numbers is 4 or less
This matches (III) 2/15.
The outcomes are (1, 2), (2, 1), (1, 3), (3, 1); the doubles (1, 1) and (2, 2) are excluded:
\( \displaystyle P = \frac{4}{30} = \frac{2}{15} \)
(C) The sum of the numbers is 4
This matches (II) 1/15.
The outcomes are (1, 3), (3, 1); the double (2, 2) is excluded:
\( \displaystyle P = \frac{2}{30} = \frac{1}{15} \)
(D) The sum of the numbers is 7
This matches (IV) 3/15.
The outcomes are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3):
\( \displaystyle P = \frac{6}{30} = \frac{3}{15} \)
So the matching is (A)-(I), (B)-(III), (C)-(II), (D)-(IV).
Note: It seems that there is a mistake in the original NTA paper, which prints the first sentence of the question incompletely: words are missing after “the probability that”, and the sentence runs straight into “Match List-I with List-II”. The question can still be solved, because the meaning is clear: each event in List-I is to be matched with its probability, given that the two numbers are different.