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Probability basic question

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Intern
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Joined: 29 May 2017
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Probability basic question  [#permalink]

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New post 17 Jul 2018, 06:04
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

the correct answer is 20/77.

can anyone explain what is wrong with my approach?
Manager
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Joined: 09 Jun 2014
Posts: 56
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
GMAT 1: 720 Q50 V38
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Probability basic question  [#permalink]

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New post 17 Jul 2018, 07:20
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

the correct answer is 20/77.

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!!
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Re: Probability basic question  [#permalink]

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New post 17 Jul 2018, 07:28
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
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Joined: 09 Jun 2014
Posts: 56
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Concentration: General Management, Operations
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Re: Probability basic question  [#permalink]

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New post 17 Jul 2018, 07:31
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, 07:31
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