Math, asked by pacrat3038, 1 year ago

A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is

Answers

Answered by chandrabhupesh
1
hey bdy....
its urr ans.....

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).
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