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Tough and tricky 6: Probability of drawing

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Expert Post
Math Expert
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Kudos [?]: 47458 [0], given: 7123

Tough and tricky 6: Probability of drawing [#permalink] New post 11 Oct 2009, 17:17
Expert's post
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

71% (01:57) correct 29% (00:36) wrong based on 30 sessions
Senior Manager
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Kudos [?]: 215 [0], given: 13

Re: Tough and tricky 6: Probability of drawing [#permalink] New post 11 Oct 2009, 17:45
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

\((3C1*2C1)*2C1/9C2 = 1/3\)
Answer is C.
Manager
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Re: Tough and tricky 6: Probability of drawing [#permalink] New post 11 Oct 2009, 22:57
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] New post 12 Oct 2009, 03:02
answer should be D

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

Ans = 3*2*2/(9*9)
Expert Post
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Kudos [?]: 47458 [0], given: 7123

Re: Tough and tricky 6: Probability of drawing [#permalink] New post 12 Oct 2009, 08:45
Expert's post
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Re: Tough and tricky 6: Probability of drawing [#permalink] New post 11 Feb 2011, 03:43

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, 03:43
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