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# There's a basket containing 10 apples - 7 red and 3 green.

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Senior Manager
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There's a basket containing 10 apples - 7 red and 3 green. [#permalink]  30 May 2006, 00:55
There's a basket containing 10 apples - 7 red and 3 green. What's the probability that upon drawing 3 apples, we get 2 red and 1 green?

I am sorry it'd take me a while to get the OA (I encountered this question in one of the GMATPrep tests I believe - but would have to look back to figure the source) but I'd appreciate if someone would post the entire methodology to solve this problem.

I guess the answer would differ if the apples are drawn 3 in one go, or drawn one after another, without replacement. Can someone provide the solutions to both approaches?

Thanks,

Kapslock
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Re: Question on Probability [#permalink]  30 May 2006, 01:18
kapslock wrote:
There's a basket containing 10 apples - 7 red and 3 green. What's the probability that upon drawing 3 apples, we get 2 red and 1 green?

I am sorry it'd take me a while to get the OA (I encountered this question in one of the GMATPrep tests I believe - but would have to look back to figure the source) but I'd appreciate if someone would post the entire methodology to solve this problem.

I guess the answer would differ if the apples are drawn 3 in one go, or drawn one after another, without replacement. Can someone provide the solutions to both approaches?

Thanks,
Kapslock

I guess soln is (7C2*3C1)/10C3

Methodology similar to
http://www.gmatclub.com/phpbb/viewtopic ... ht=#201859
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Method 1

Two red and 1 green can be arranged in
RRG
RGR
GRR
ways
RRG = 7/10 * 6/9 * 3/8 = 7/40
All the three ways have the probability of 7/40
Hence total probabiltiy = 21/40

Method 2

7C2*3C1/10C3 = 21/40
Senior Manager
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Posts: 419
Location: Phoenix
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Kudos [?]: 12 [0], given: 0

Thanks shobhitb and jaynayak !!!

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