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

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

02 Apr 2008, 10:28

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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?

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Re: PS probability--apples [#permalink]

02 Apr 2008, 10:34

There are three different ways that the apples can be picked (RRG, RGR, GRR)

Let's look at the first case (RRG)

P(RRG) = 7/10*6/9*3/8

Since we said the order of picking apples does not matter, we will have to similarly calculate probability for RGR and GRR cases which works out to the same.

So P(2Red,1Green) = 3*(7/10*6/9*3/8) = 21/40

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Re: PS probability--apples [#permalink]

02 Apr 2008, 11:51

\frac{C_7^2*C_3^1}{C_{10}^3} = \frac{21}{40}
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Re: PS probability--apples [#permalink]

02 Apr 2008, 12:32

neelesh wrote:

\frac{C_7^2*C_3^1}{C_{10}^3} = \frac{21}{40}

good, agreed...can you easily solve a related problem: 7-t62090

Re: PS probability--apples [#permalink] 02 Apr 2008, 12:32

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

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

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

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