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) | \(\displaystyle\frac{12}{35}\) |
| (B) | Only 1 red ball is | (II) | \(\displaystyle\frac{1}{35}\) |
| (C) | Exactly 2 red balls is | (III) | \(\displaystyle\frac{4}{35}\) |
| (D) | no white ball is | (IV) | \(\displaystyle\frac{18}{35}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Three balls are drawn from 7, so the number of equally likely selections is
\({}^{7}C_{3} = 35\)
Each probability is a count of selections divided by 35.
(A) No red ball
This matches (III) \(\frac{4}{35}\): all three are white, \({}^{4}C_{3} = 4\).
(B) Only 1 red ball
This matches (IV) \(\frac{18}{35}\): one red and two white,
\({}^{3}C_{1} \times {}^{4}C_{2} = 3 \times 6 = 18\)
(C) Exactly 2 red balls
This matches (I) \(\frac{12}{35}\): two red and one white,
\({}^{3}C_{2} \times {}^{4}C_{1} = 3 \times 4 = 12\)
(D) no white ball
This matches (II) \(\frac{1}{35}\): all three are red, \({}^{3}C_{3} = 1\).
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II). As a check, 4 + 18 + 12 + 1 = 35.