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#1
08-24-2007, 03:25 PM
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A probability question.
A set of ten objects, in which two people are supposed to choose six out of the ten, at different times, and with no knowledge of the other person's picks.
What are the chances that the two people will pick the exact same six objects? What are the chances of having five in common? How do these probabilities change if a third person is included? 
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#2
08-24-2007, 04:55 PM
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Re: A probability question.
For the first question. Person A picks his 6, it really doesnt matter which. Then person B has the following chance to pick the same:
6/10 * 5/9 * 4/8 * 3/7 * 2/6 * 1/5= 720/151200 = 1/210 If you add a 3rd person they have a 1/210 chance to pick the same, so 1/210 * 1/210 = 1/44100
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#3
08-24-2007, 05:49 PM
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Re: A probability question.
Daddy is right, for 2 people picking exactly the same 6 objects. 1 way out of 10C6=210.
5 objects in common: 6C5*4C1 = 6*4 = 24 ways 4 objects in common: 6C4*4C2 = 15*6 = 90 ways 3 objects in common: 6C3*4C3 = 20*4 = 80 ways 2 objects in common: 6C2*4C4 = 15*1 = 15 ways Impossible to have fewer than two objects in common.
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#4
08-25-2007, 12:19 PM
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Re: A probability question.
Another way of thinking about this problem is to paint six of the objects red. One person picks six objects at random, the chance of picking k red objects is:
[6! 4! / (6 - n)!]^2 / [10! n! (n-2)!], 6 >= n > 1 This is clearly the same as two people picking n objects in common. With more people, you just multiply this by this number, as Daddys_Visa pointed out.
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