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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
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
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04 Nov 2005, 19:58
Kudos
Bookmarks
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 .......
RK
can you pls explanin "calculate the hypergeometric dist. probability with replacement " .....I don't know what exactly this means .....
total no of ways to get 6 picture cards 12C6*40C7
Probability 12C6*40C6/52C13
Am i right .......
RK
can you pls explanin "calculate the hypergeometric dist. probability with replacement " .....I don't know what exactly this means .....
04 Nov 2005, 20:04
Kudos
Bookmarks
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?
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?
