Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.
Match List-I with List-II
| List-I | List-II | ||
| X | Probability, P(X) | ||
| (A) | 4 | (I) | \(\displaystyle\frac{1}{6}\) |
| (B) | 5 | (II) | \(\displaystyle\frac{5}{36}\) |
| (C) | 6 | (III) | \(\displaystyle\frac{1}{12}\) |
| (D) | 7 | (IV) | \(\displaystyle\frac{1}{9}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Two dice give 36 equally likely outcomes, so P(X) is the number of ordered pairs with sum X, divided by 36.
(A) X = 4: (1, 3), (2, 2), (3, 1), so 3 outcomes.
\( \displaystyle P(X = 4) = \frac{3}{36} = \frac{1}{12} \)
This matches (III).
(B) X = 5: (1, 4), (2, 3), (3, 2), (4, 1), so 4 outcomes.
\( \displaystyle P(X = 5) = \frac{4}{36} = \frac{1}{9} \)
This matches (IV).
(C) X = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), so 5 outcomes.
\( \displaystyle P(X = 6) = \frac{5}{36} \)
This matches (II).
(D) X = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), so 6 outcomes.
\( \displaystyle P(X = 7) = \frac{6}{36} = \frac{1}{6} \)
This matches (I).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).