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# Probability

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Manager
Joined: 08 Feb 2009
Posts: 144
Schools: Anderson

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02 May 2009, 09:55
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A jar contains only red, yellow, and orange marbles. If there are 3 red, 5 yellow, and 4 orange
marbles, and 3 marbles are chosen from the jar at random without replacing any of them, what
is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
Manager
Joined: 12 Apr 2006
Posts: 213
Location: India

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02 May 2009, 20:08
1
KUDOS
This is what I got

Red 3
Yellow 5
Orange 4

In this we have to choose 1 red and 2 yellow marbles.
You can choose marbles in 3 different ways

R, Y, Y
$$3/12 * 5/11 * 4/10$$

Y, R, Y
$$5/12 * 3/11 * 4/10$$

Y, Y, R

$$5/12 * 4/11 * 3/10$$

So it comes to
$$3*(3*4*5/10*11*12) = 3/22$$ Ans

What is the OA?
Manager
Joined: 08 Feb 2009
Posts: 144
Schools: Anderson

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03 May 2009, 04:44
humans wrote:
This is what I got

Red 3
Yellow 5
Orange 4

In this we have to choose 1 red and 2 yellow marbles.
You can choose marbles in 3 different ways

R, Y, Y
$$3/12 * 5/11 * 4/10$$

Y, R, Y
$$5/12 * 3/11 * 4/10$$

Y, Y, R

$$5/12 * 4/11 * 3/10$$

So it comes to
$$3*(3*4*5/10*11*12) = 3/22$$ Ans

What is the OA?

3/22 is the OA.
Thanks for that.
Manager
Joined: 10 May 2009
Posts: 65

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18 May 2009, 06:28
I got the correct answer but is this the correct approach:

(5C2*3C1)/12C3=3/22
Manager
Joined: 12 Apr 2006
Posts: 213
Location: India

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19 May 2009, 02:37
prinits wrote:
I got the correct answer but is this the correct approach:

(5C2*3C1)/12C3=3/22

Yes this is correct approach, rather I say formula based approach. Sometimes I get confused where to use permuatation and combination formulas so use the above described method from Manhattan math book.
Re: Probability   [#permalink] 19 May 2009, 02:37
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