Total number of balls = 10 + 15 = 25 (a)
(a) Let event A: First ball drawn is red.
∴ P(A) =\(\frac {^{10}C_1}{^{25}C_1} = \frac {10} {25} = \frac {2}{5}\)
Let event B: Second ball drawn is green.
Since the first red ball is not replaced in the box, we now have 24 balls, out of which 15 are green.
∴ Probability that the second ball is green under the condition that the first red ball is not replaced in the box = P(B/A) =
\(\frac {^{15}C_1}{^{24}C_1} = \frac {15} {24} = \frac {8}{5}\)
∴ Required probability = P(A ∩ B) = P(B/A) . P(A)
= 2/3 x 5/8
= 1/4
(b) To find the probability that one ball is red and the other is green, there are two possibilities:
First ball is red and second ball is green.
OR
The first ball is the green and the second ball is red.
From above, we get
P(First ball is red and second ball is green) = 1/4
Similarly,
P(First ball is green and second ball is red) =
\(\frac {^{15}C_1}{^{24}C_1}\)x \(\frac {^{10}C_1}{^{24}C_1} = \frac {15} {25} \) x \(\frac {10} {24} = \frac {1}{4}\)
∴ Required probability = P(First ball is red and second ball is green) + P(First ball is green and second ball is red)
= 1/4 + 1/4
= 1/2
