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Fabian
09-28-2005, 08:11 PM
Hi guys I'm sorry for not knowing how to calculate this.

You have a deck of 60 cards. 39 are red and 21 are black. You draw 26 cards. What are the chances at least 13 of them are black?

Thanks.

pzhon
09-28-2005, 11:25 PM
[ QUOTE ]

You have a deck of 60 cards. 39 are red and 21 are black. You draw 26 cards. What are the chances at least 13 of them are black?

[/ QUOTE ]
There are 60 choose 26 (http://www.google.com/search?hl=en&amp;lr=&amp;q=60+choose+26) equally likely ways to draw 26 cards.

The number of ways to draw exactly n black cards and 26-n red cards is (21 choose n) * (39 choose 26-n).

Sum from n=13 to 26 (or from n=13 to 21, since the other terms are 0) to get the total number of ways of having at least 13 black cards. Divide by 60 choose 26 to get the probability.

2212819605429930/69886166503903470

~0.0316632

Luzion
09-29-2005, 12:56 AM
[ QUOTE ]
[ QUOTE ]

You have a deck of 60 cards. 39 are red and 21 are black. You draw 26 cards. What are the chances at least 13 of them are black?

[/ QUOTE ]
There are 60 choose 26 (http://www.google.com/search?hl=en&amp;lr=&amp;q=60+choose+26) equally likely ways to draw 26 cards.

The number of ways to draw exactly n black cards and 26-n red cards is (21 choose n) * (39 choose 26-n).

Sum from n=13 to 26 (or from n=13 to 21, since the other terms are 0) to get the total number of ways of having at least 13 black cards. Divide by 60 choose 26 to get the probability.

2212819605429930/69886166503903470

~0.0316632

[/ QUOTE ]

Just wanted to clarify a little on pzhon's nice answer. Its a hypergeometric question so the formula for this problem would be.

[ C(21,n) * C(39,26-n) ] / C(60,26)

Sum up all the values for n = (13,14,15,16,17,18,19,20,21)

Fabian
09-29-2005, 12:38 PM
Thanks alot for the answers pzhon and Luzion.

peterchi
09-30-2005, 09:27 AM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

You have a deck of 60 cards. 39 are red and 21 are black. You draw 26 cards. What are the chances at least 13 of them are black?

[/ QUOTE ]
There are 60 choose 26 (http://www.google.com/search?hl=en&amp;lr=&amp;q=60+choose+26) equally likely ways to draw 26 cards.

The number of ways to draw exactly n black cards and 26-n red cards is (21 choose n) * (39 choose 26-n).

Sum from n=13 to 26 (or from n=13 to 21, since the other terms are 0) to get the total number of ways of having at least 13 black cards. Divide by 60 choose 26 to get the probability.

2212819605429930/69886166503903470

~0.0316632

[/ QUOTE ]

Just wanted to clarify a little on pzhon's nice answer. Its a hypergeometric question so the formula for this problem would be.

[ C(21,n) * C(39,26-n) ] / C(60,26)

Sum up all the values for n = (13,14,15,16,17,18,19,20,21)

[/ QUOTE ]
Hey Luzion (or anyone):

Can you give me some intuition behind the hypergeometric distribution?

It was on our last homework assignment in my probability class, and our professor never mentioned it in lecture, nor was it in any of our texts. I got the formula from a friend, did a plug-and-chug, and got the question right... but I don't really have any true understanding of why it works.

Thanks

LetYouDown
09-30-2005, 10:26 AM
Hypergeometric Distribution (http://mathworld.wolfram.com/HypergeometricDistribution.html)