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# Probability basic question

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Manager
Joined: 29 May 2017
Posts: 100
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability

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17 Jul 2018, 05:04
There are 4 red, 5 blue and 2 green balls in a bag. If 5 balls are selected one after the other without replacement, what is the probability of
choosing 2 red, 2 blue and 1 green ball?

My approach to this is as follows:

1. find the probability of ONE SUCH EVENT: p(2r, 2b, 1g)
2. multiply the above with the number of ways the selections can be made: this is simply: C(4,2) x C(5,2) x C(2,1)=120

as for 1:

lets say p(GRBRB)= 2/11 x 4/10 x 5/9 x 3/8 x 4/7 = 2/11 x 1/3 x 1/7

thus giving 120 x 2/11 x 1/3 x 1/7 = 80/77

can anyone explain what is wrong with my approach?
Manager
Joined: 09 Jun 2014
Posts: 190
Location: India
Concentration: General Management, Operations
Schools: Tuck '19

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17 Jul 2018, 06:20
Mansoor50 wrote:
There are 4 red, 5 blue and 2 green balls in a bag. If 5 balls are selected one after the other without replacement, what is the probability of
choosing 2 red, 2 blue and 1 green ball?

My approach to this is as follows:

1. find the probability of ONE SUCH EVENT: p(2r, 2b, 1g)
2. multiply the above with the number of ways the selections can be made: this is simply: C(4,2) x C(5,2) x C(2,1)=120

as for 1:

lets say p(GRBRB)= 2/11 x 4/10 x 5/9 x 3/8 x 4/7 = 2/11 x 1/3 x 1/7

thus giving 120 x 2/11 x 1/3 x 1/7 = 80/77

can anyone explain what is wrong with my approach?

I think in your case since there are 2R and 2B selections .And because these are equivalent so possible combinations needs to be divided by 2*2(Since if p objects are alike and one kind..)

Basic selection give the correct answer here.

I followed the below approach.

(4C2*5C2*2C1)/11C5

Press Kudos if it helps!!
Intern
Joined: 25 Apr 2018
Posts: 1

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17 Jul 2018, 06:28
In P(GRBRB) since the position of 2 red and 2 green if interchanged does not make any effect. So it should be 4C2/2 and 5C2/2

Posted from my mobile device
Manager
Joined: 09 Jun 2014
Posts: 190
Location: India
Concentration: General Management, Operations
Schools: Tuck '19

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17 Jul 2018, 06:31
ShanmukhaMD wrote:
In P(GRBRB) since the position of 2 red and 2 green if interchanged does not make any effect. So it should be 4C2/2 and 5C2/2

Posted from my mobile device

Exactly!!

Hope it helps.
Re: Probability basic question &nbs [#permalink] 17 Jul 2018, 06:31
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