A bag contains 6 red and 4 blue balls. Two balls are drawn at random, one after the other, without replacement.
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Probability that both balls are red | (I) | \(\displaystyle\frac{2}{15}\) |
| (B) | Probability that both balls are blue | (II) | \(\displaystyle\frac{2}{3}\) |
| (C) | Probability that one ball is red and the other is blue | (III) | \(\displaystyle\frac{1}{3}\) |
| (D) | Probability that the second ball is red, given that the first ball is blue | (IV) | \(\displaystyle\frac{8}{15}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Without replacement, the bag has 9 balls left for the second draw. Each case uses the multiplication theorem: P(first) × P(second | first).
(A) Probability that both balls are red
This matches (III) \(\frac{1}{3}\).
\( \displaystyle P = \frac{6}{10}\times\frac{5}{9} = \frac{30}{90} = \frac{1}{3} \)
(B) Probability that both balls are blue
This matches (I) \(\frac{2}{15}\).
\( \displaystyle P = \frac{4}{10}\times\frac{3}{9} = \frac{12}{90} = \frac{2}{15} \)
(C) Probability that one ball is red and the other is blue
This matches (IV) \(\frac{8}{15}\).
The red ball may come first or second, and these two orders are added:
\( \displaystyle P = \frac{6}{10}\times\frac{4}{9}+\frac{4}{10}\times\frac{6}{9} = \frac{48}{90} = \frac{8}{15} \)
(D) Probability that the second ball is red, given that the first ball is blue
This matches (II) \(\frac{2}{3}\).
After a blue ball is taken out, 9 balls are left, of which 6 are red:
\( \displaystyle P = \frac{6}{9} = \frac{2}{3} \)