The bag contains 5 red and 4 black balls, i.e., 5 + 4 = 9 balls.
(i) 2 balls can be drawn from 9 balls with replacement in 9C1 X 9C1 ways.
∴ n(S) = 9C1 X 9C1 = 9 × 9 = 81
Let event A: Balls drawn are red. 2 red balls can be drawn from 5 red balls with replacement in 5C1X 5C1 ways.
∴ n(A) = 5C1X 5C1 = 5 × 5 = 25
∴ P(A) = \(\frac {n(A)} {n (S)} = \frac {25}{81}\)
(ii) 2 balls can be drawn from 9 balls without replacement in 9C1X 8C1 ways.
∴ n(S) = 9C1X 8C1= 9 × 8 = 72
2 red balls can be drawn from 5 red balls without replacement in
5C1X 4C1 ways.
∴ n(B) = 5C1X 4C1= 5 × 4 = 20
∴ P(B) = \(\frac {n(B)} {n (S)} = \frac {20}{72} = \frac 5{18}\)