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# In an urn, there are 2 red balls, 3 blue balls, and 3 green

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Manager
Joined: 04 Nov 2006
Posts: 158
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Kudos [?]: 8 [0], given: 0

In an urn, there are 2 red balls, 3 blue balls, and 3 green [#permalink]  01 Dec 2006, 18:39
In an urn, there are 2 red balls, 3 blue balls, and 3 green balls. If five balls are picked, without replacement, what is the probability that 1 red ball, 2 green balls, and 2 blue balls will be picked?

Can anyone provide a method for this problem?
Intern
Joined: 24 Feb 2006
Posts: 48
Followers: 1

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

I 'm getting 3/280
whats the OA?
Director
Joined: 28 Dec 2005
Posts: 922
Followers: 2

Kudos [?]: 37 [0], given: 0

Lemme give this a try. Probability as usual always a problem for me.

The total number of balls = 8
Number of ways of choosing 5 balls from 8 without replacement: 8C5

Number of ways of choosing 1 red from 2 red balls: 2C1
Number of ways of choosing 2 blue from 3 blueballs:3C2
Number of ways of choosing 2 green from 3 green balls:3C2

=> total number of ways of choosing this set: 2C1*3C2*3C2

=> probability: 2C1*3C2*3C2/8C5 = 9/28

Let me know if you think Ive done something wrong
Manager
Joined: 04 Nov 2006
Posts: 158
Followers: 3

Kudos [?]: 8 [0], given: 0

Nice job, Sampath. That is the correct answer. I had no idea on this one when I saw it on a practice test earlier.
Manager
Joined: 25 Nov 2006
Posts: 60
Followers: 2

Kudos [?]: 2 [0], given: 0

selecting 1 out of 2 red balls can be done in 2C1 ways
selecting 2 out of 3 Blue balls can be done in 3C2 ways
selecting 2 out of 3 Green balls can be done in 3C2 ways

The selecting of 5 balls out of 8 can be done in 8C5 ways.
hence the probability is 9/28.
Manager
Joined: 04 Nov 2006
Posts: 158
Followers: 3

Kudos [?]: 8 [0], given: 0

As always, tennis, thanks for that detailed explanation
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