Question

A box a contain of 2 white and 4 black balls another box b contain 5 white and 7 black a ball is transfered

Answer 1

The bag contains 2 White, 3 Black and 4 Red balls.

So, total 9 balls are there in the bag; among them 3 are Black and 6 are non-Black balls.

Three balls can randomly be drawn in (9C3) = 84 ways.

1 Black and 2 non-Black balls can be drawn in (3C1)*(6C2) = 45 ways.

1 non-Black and 2 Black balls can be drawn in (6C1)*(3C2) = 18 ways.

3 non-Black balls can be drawn in (3C3) = 1 way.

So, three balls drawn in (45 + 18 + 1) = 64 ways will have at least one Black ball among the drawn ones.

So, the probability of getting at least one Black ball among the three drawn =

(64 / 84) = (16 / 21).