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) $\frac{1}{6}$
(B) 5	(II) $\frac{5}{36}$
(C) 6	(III) $\frac{1}{12}$
(D) 7	(IV) $\frac{1}{9}$

Choose the correct answer from the options given below:

When two dice are thrown, each die can show 1, 2, 3, 4, 5, or 6.

Total possible outcomes $= 6 \times 6 = 36$

$P(X) = \frac{\text{Number of ways to get sum X}}{36}$

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For X = 4, the combinations are:

• (1, 3)
• (2, 2)
• (3, 1)

Total ways $= 3$

$P(X=4) = \frac{3}{36}$

$P(X=4) = \frac{1}{12}$

This matches with (III)

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For X = 5, the combinations are:

• (1, 4)
• (2, 3)
• (3, 2)
• (4, 1)

Total ways $= 4$

$P(X=5) = \frac{4}{36}$

$P(X=5) = \frac{1}{9}$

This matches with (IV)

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For X = 6, the combinations are:

• (1, 5)
• (2, 4)
• (3, 3)
• (4, 2)
• (5, 1)

Total ways $= 5$

$P(X=6) = \frac{5}{36}$

This matches with (II)

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For X = 7, the combinations are:

• (1, 6)
• (2, 5)
• (3, 4)
• (4, 3)
• (5, 2)
• (6, 1)

Total ways $= 6$

$P(X=7) = \frac{6}{36}$

$P(X=7) = \frac{1}{6}$

This matches with (I)

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The matching is:

• (A) X = 4 → P = $\frac{1}{12}$ → (III)
• (B) X = 5 → P = $\frac{1}{9}$ → (IV)
• (C) X = 6 → P = $\frac{5}{36}$ → (II)
• (D) X = 7 → P = $\frac{1}{6}$ → (I)