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
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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).
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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