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Hypergeometric Distribution Question

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Senior Manager
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Hypergeometric Distribution Question [#permalink] New post 04 Nov 2005, 18:31
Hi All,

I am aware of a method to calculate the hypergeometric dist. probability with replacement . What is the method to calculate the same without replacement.

I got a question: If 13 cards are selected without replacement from deck of 52 find the probability that 6 will be picture card.

Actual ans not required, but please post the equation of the ans.

Thanks

Last edited by hkm_gmat on 04 Nov 2005, 19:52, edited 1 time in total.
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 [#permalink] New post 04 Nov 2005, 18:58
tatal no of ways to get 13 cards 52C13

total no of ways to get 6 picture cards 12C6*40C7

Probability 12C6*40C6/52C13

Am i right .......:roll:


RK

can you pls explanin "calculate the hypergeometric dist. probability with replacement " .....I don't know what exactly this means .....:roll:

8-)
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 [#permalink] New post 04 Nov 2005, 19:04
I'm assuming non replacement


All 5 preceding cards Picture:

(40!/40!)(12!/6!)*5C0/(52!/46!)


One of the 5 preceding cards are non picture

(40!/39!)(12!/7!)*5C1/(52!/46!)


Two of the 5 preceding cards are non picture

(40!/38!)(12!/8!)*5C2/(52!/46!)


Three of the 5 preceding cards are non picture

(40!/37!)(12!/9!)*5C3/(52!/46!)


Four of the 5 preceding cards are non picture

(40!/36!)(12!/10!)*5C4/(52!/46!)

Five of the 5 preceding cards are non picture

(40!/35!)(12!/11!)*5C5/(52!/46!)


just add em up! easy right?


:wink:
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 [#permalink] New post 04 Nov 2005, 19:55
cool_jonny009 wrote:
tatal no of ways to get 13 cards 52C13

total no of ways to get 6 picture cards 12C6*40C7

Probability 12C6*40C6/52C13

Am i right .......:roll:


RK

can you pls explanin "calculate the hypergeometric dist. probability with replacement " .....I don't know what exactly this means .....:roll:

8-)



Hi , Thats not it. This question is based on "without replacement". We need to find the probability without replacement.

I am sorry I was not clear the first time around. The method used by you is with replacement. With Replacement means the selected item is returned to the pool from which the cards are drawn from.
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 [#permalink] New post 04 Nov 2005, 21:58
hkm_gmat wrote:
Any more takers ??


Dude I wouldn't lose sleep over this one. Not likely to be asked on the actual GMAT. Probability problems from even the difficult bin are quite straight forward.
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 [#permalink] New post 06 Nov 2005, 07:09
hkm_gmat wrote:
cool_jonny009 wrote:
tatal no of ways to get 13 cards 52C13

total no of ways to get 6 picture cards 12C6*40C7

Probability 12C6*40C6/52C13

Am i right .......:roll:


RK

can you pls explanin "calculate the hypergeometric dist. probability with replacement " .....I don't know what exactly this means .....:roll:

8-)



Hi , Thats not it. This question is based on "without replacement". We need to find the probability without replacement.

I am sorry I was not clear the first time around. The method used by you is with replacement. With Replacement means the selected item is returned to the pool from which the cards are drawn from.


I don't believe that is correct. With replacement, the answer would be:
C(13,6)(12/52)^6(40/52)^7.

I think what RK has proposed is the right solution. Do you have OA?
  [#permalink] 06 Nov 2005, 07:09
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