The bag contains 7 black and 4 red balls,
i.e., 7 + 4 = 11 balls.
∴ 3 balls can be drawn out of 11 balls in 11C3 ways.
∴ n(S) = 11C3
(i) Let event A: All 3 balls drawn are black.
There are 7 black balls.
∴ 3 black balls can be drawn from 7 black balls in 7C3 ways.
∴ n(A) = 7C3
(ii) Let event B: Out of 3 balls drawn, one is black and two are red.
There are 7 black and 4 red balls.
∴ One black ball can be drawn from 7 black balls in 7C1 ways and 2 red balls can be drawn from 4 red balls in 4C2 ways.
∴ n(A) = 7C1 x 4C2
