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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
|
List-I |
List-II |
|
(A) No red ball is |
(III) $\frac{4}{35}$ |
|
(B) Only 1 red ball is |
(IV) $\frac{18}{35}$ |
|
(C) Exactly 2 red balls is |
(I) $\frac{12}{35}$ |
|
(D) no white ball is |
(II) $\frac{1}{35}$ |
Total balls = 4 white + 3 red = 7
Total ways to choose 3 balls = 7C3 = 35
(A) No red ball ⇒ all 3 white
Ways = 4C3 = 4
Probability = \(\frac{4}{35}\) ⇒ (A) → (III)
(B) Only 1 red ball ⇒ 1 red, 2 white
Ways = 3C1 × 4C2 = 3 × 6 = 18
Probability = \(\frac{18}{35}\) ⇒ (B) → (IV)
(C) Exactly 2 red balls ⇒ 2 red, 1 white
Ways = 3C2 × 4C1 = 3 × 4 = 12
Probability = \(\frac{12}{35}\) ⇒ (C) → (I)
(D) No white ball ⇒ all 3 red
Ways = 3C3 = 1
Probability = \(\frac{1}{35}\) ⇒ (D) → (II)