A box contains 4 red balls and 2 white balls. Balls are drawn at random from the box one by one, without replacement. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Probability that the first white ball is obtained on the first draw | (I) | \(\displaystyle\frac{1}{5}\) |
| (B) | Probability that the first white ball is obtained on the second draw | (II) | \(\displaystyle\frac{1}{15}\) |
| (C) | Probability that the first white ball is obtained on the third draw | (III) | \(\displaystyle\frac{1}{3}\) |
| (D) | Probability that both white balls are obtained in the first two draws | (IV) | \(\displaystyle\frac{4}{15}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each draw is made from the balls left after the earlier draws, so the multiplication theorem is used with the changing numbers of balls.
(A) Probability that the first white ball is obtained on the first draw
This matches (III) \(\frac{1}{3}\).
\( \displaystyle P = \frac{2}{6} = \frac{1}{3} \)
(B) Probability that the first white ball is obtained on the second draw
This matches (IV) \(\frac{4}{15}\).
The first ball is red and the second is white:
\( \displaystyle P = \frac{4}{6}\times\frac{2}{5} = \frac{4}{15} \)
(C) Probability that the first white ball is obtained on the third draw
This matches (I) \(\frac{1}{5}\).
The first two balls are red and the third is white:
\( \displaystyle P = \frac{4}{6}\times\frac{3}{5}\times\frac{2}{4} = \frac{1}{5} \)
(D) Probability that both white balls are obtained in the first two draws
This matches (II) \(\frac{1}{15}\).
The first ball is white and the second is the other white ball:
\( \displaystyle P = \frac{2}{6}\times\frac{1}{5} = \frac{1}{15} \)