Probability question

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bhavna: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Sat Apr 29, 2006 4:17 pm

by bhavna » Sun May 28, 2006 12:11 pm

Richard has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 without replacement, what is the probability that of the 5 drawn balls he has picked 1 red, 2 green and 2 blue balls?

The explanation gives the answer as 9/28.

My answer was the following:

(2C1 * 6C4)/8C5 * (3C2 * 5C3)/8C5 * (3C2 * 5C3)/8C5

Can you please explain how we got 9/28?

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Source: Beat The GMAT — Problem Solving | Sun May 28, 2006 12:11 pm

rehanaj: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Tue Jul 18, 2006 1:23 pm

by rehanaj » Tue Jul 18, 2006 1:28 pm

The probability to get the exact sequence of 1R-2G-2B is

[C(2,1)/C(8,1)] x [C(3,2)/C(7,2)] x [C(3,2)/C(5,2)] = 3/280

However, we can get 1R-2G-2B in many different ways (because the order does not matter). More specifically, we can get 1R-2G-2B in

C(5,1) x C(4,2) x C(2,2) = 30 ways

Required probability is thus

3/280 x 30 = 9/28

Rehana

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members; Thanked: 11 times

by 800guy » Tue Jul 18, 2006 5:28 pm

rehanaj wrote:The probability to get the exact sequence of 1R-2G-2B is

[C(2,1)/C(8,1)] x [C(3,2)/C(7,2)] x [C(3,2)/C(5,2)] = 3/280

However, we can get 1R-2G-2B in many different ways (because the order does not matter). More specifically, we can get 1R-2G-2B in

C(5,1) x C(4,2) x C(2,2) = 30 ways

Required probability is thus

3/280 x 30 = 9/28

Rehana

thanks for the excellent explantion!!

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Thouraya: Master | Next Rank: 500 Posts; Posts: 112; Joined: Wed Jan 20, 2010 5:46 am; Thanked: 1 times

by Thouraya » Tue Jun 01, 2010 4:57 am

Hey Rehana,

Im sorry Im aware that ur explanation is very detailed, but honestly I didn't understand the solution..I know ur using the combinations formula, but why are u dividing the combinations?and how did u come up with the 30?Thank u!

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Tue Jun 01, 2010 5:53 am

Another apprioach for this problem

Prob of selecting 1R,2G and 2b= Prob of favourable outcomes/Prob of total number of outcomes

Prob of favourable outcomes(1R,2G,2B)=2C1*3C2*3C2

Prob of total number of outcomes(selecting 5 balls from 8 balls)=8C5

=2C1*3C2*3C2/8C5=9/28

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Gmat Bond: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Sat Mar 06, 2010 12:13 pm; Location: Secret

by Gmat Bond » Tue Jun 01, 2010 8:04 am

Good one.

The name is Bond, GMAT BOND !!

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