Match List-I with List-II

An urn contains 4 white and 3 red balls. In a random draw of three balls, the probability of

List-I	List-II
(A) No red ball is	(I) $\frac{12}{35}$
(B) Only 1 red ball is	(II) $\frac{1}{35}$
(C) Exactly 2 red balls is	(III) $\frac{4}{35}$
(D) no white ball is	(IV) $\frac{18}{35}$

Choose the correct answer from the options given below:

The urn contains 4 white balls and 3 red balls, giving a total of 7 balls. Three balls are drawn randomly.

The total number of ways to choose 3 balls from 7:

$C(7,3) = \frac{7!}{3! \times 4!}$

$= \frac{7 \times 6 \times 5}{3 \times 2 \times 1}$

$= 35$

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(A) No red ball means all 3 balls are white.

Number of ways to choose 3 white balls from 4 white balls:

$C(4,3) = 4$

Probability $= \frac{4}{35}$

Match: (A) → (III)

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(B) Only 1 red ball means 1 red and 2 white balls.

Number of ways to choose 1 red from 3 red balls: $C(3,1) = 3$

Number of ways to choose 2 white from 4 white balls: $C(4,2) = 6$

Total ways $= 3 \times 6 = 18$

Probability $= \frac{18}{35}$

Match: (B) → (IV)

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(C) Exactly 2 red balls means 2 red and 1 white ball.

Number of ways to choose 2 red from 3 red balls: $C(3,2) = 3$

Number of ways to choose 1 white from 4 white balls: $C(4,1) = 4$

Total ways $= 3 \times 4 = 12$

Probability $= \frac{12}{35}$

Match: (C) → (I)

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(D) No white ball means all 3 balls are red.

Number of ways to choose 3 red balls from 3 red balls:

$C(3,3) = 1$

Probability $= \frac{1}{35}$

Match: (D) → (II)

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The correct matching is:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)