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# Tough and tricky 6: Probability of drawing

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Tough and tricky 6: Probability of drawing [#permalink]

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11 Oct 2009, 18:17
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A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
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Re: Tough and tricky 6: Probability of drawing [#permalink]

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11 Oct 2009, 18:45
Number of ways to choose Red ball = 3C1
Number of ways to choose White ball = 2C1
Total number of ways to choose 2 balls at random = 9C2

$$(3C1*2C1)*2C1/9C2 = 1/3$$
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Re: Tough and tricky 6: Probability of drawing [#permalink]

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11 Oct 2009, 23:57
I think its D i.e. 4/27

we can have either RW or WR combination in successive draws.

So the probability will be (3/9 x 2/9) + (2/9 x 3/9) = 4/27

OA pls???????
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Re: Tough and tricky 6: Probability of drawing [#permalink]

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12 Oct 2009, 04:02

As we are replacing the balls so the probability of drawing two balls at random is 9*9

Ans = 3*2*2/(9*9)
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Re: Tough and tricky 6: Probability of drawing [#permalink]

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12 Oct 2009, 09:45
Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
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Re: Tough and tricky 6: Probability of drawing [#permalink]

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11 Feb 2011, 04:43

probability of getting RW = $$\frac{3}{9}*\frac{2}{9}$$
we could have RW or WR. in both cases, we'll have a red and a white ball.

total probability = $$\frac{3}{9}*\frac{2}{9}*2 = \frac{4}{27}$$

Ans: D

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Re: Tough and tricky 6: Probability of drawing   [#permalink] 11 Feb 2011, 04:43
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