My Math Forum Probability of 4 of 6 dice containing exactly one pair

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 February 7th, 2010, 05:18 AM #1 Member   Joined: Jan 2010 Posts: 32 Thanks: 0 Probability of 4 of 6 dice containing exactly one pair I am not good at solving probability problems and would appreciate any help with the below problem... You toss 6 dice. If exactly one die is a 2 and exactly one die is a 5, in how many ways can the remaining 4 dice contain exactly one pair?
 February 7th, 2010, 05:44 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability of 4 of 6 dice containing exactly one pair The remaining dice may be any of 1, 3, 4, or 6. Pick one of these to represent the pair; then, pick two of the remaining three to be the last two dice. $4{3\choose 2}= 4\cdot 3 = 12$
 February 7th, 2010, 05:58 AM #3 Member   Joined: Jan 2010 Posts: 32 Thanks: 0 Re: Probability of 4 of 6 dice containing exactly one pair Thanks a ton. Is it possible to explain little bit more in detail. I did not quite understand it. Appreciate your help.
 February 7th, 2010, 08:05 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability of 4 of 6 dice containing exactly one pair The remaining dice may be any of 1, 3, 4, or 6. Two of them have to be the same: A A B C So there are four choices for A. Then (depending on your choice for A) there are three choices for B, and two choices for C. 4*3*2 = 24. But if B and C were switched, you couldn't tell the difference, so divide by two to get 12. (You are choosing two items out of three, and order isn't important. This is represented as "three choose two", written ${3\choose 2}$. The general formula for "x choose y" is $\frac{x!}{y!(x-y)!}$.)

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