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Probability

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Manager
Manager
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Joined: 08 Feb 2009
Posts: 145

Kudos [?]: 58 [0], given: 3

Schools: Anderson
Probability [#permalink]

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New post 02 May 2009, 10:55
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Please explain your answer



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?

Kudos [?]: 58 [0], given: 3

1 KUDOS received
Manager
Manager
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Joined: 12 Apr 2006
Posts: 214

Kudos [?]: 29 [1], given: 17

Location: India
Re: Probability [#permalink]

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New post 02 May 2009, 21:08
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This post received
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?

Kudos [?]: 29 [1], given: 17

Manager
Manager
User avatar
Joined: 08 Feb 2009
Posts: 145

Kudos [?]: 58 [0], given: 3

Schools: Anderson
Re: Probability [#permalink]

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New post 03 May 2009, 05:44
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.

Kudos [?]: 58 [0], given: 3

Manager
Manager
avatar
Joined: 10 May 2009
Posts: 66

Kudos [?]: 114 [0], given: 11

Re: Probability [#permalink]

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New post 18 May 2009, 07:28
I got the correct answer but is this the correct approach:

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

Kudos [?]: 114 [0], given: 11

Manager
Manager
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Joined: 12 Apr 2006
Posts: 214

Kudos [?]: 29 [0], given: 17

Location: India
Re: Probability [#permalink]

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New post 19 May 2009, 03:37
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.

Kudos [?]: 29 [0], given: 17

Re: Probability   [#permalink] 19 May 2009, 03:37
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