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Three numbers are chosen at random without replacement from {1,2,3,.....,8}. What is the probability that minimum is 3 given that their maximum is 6?
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This question is based on conditional probability..

We know : 

P(A | B)     =      P(A ∩ B) / P(B)

So    here  P(B)     =     P(maximum of 3 numbers = 6) 

So we have 5 numbers to chosse from to choose remaining 2 numbers which can be done in 5C2  =   10 ways..

Hence  P(B)    =   10 / 8C3

Now coming to P(A ∩ B) , it suggests minimum of 3 numbers should be 3 and maximum should be 6..

So only possibility is to choose the middle number which can be 4 or 5 and hence 2 ways.

Hence P(A | B)      =  ( 2/8C3 )  /  (10 / 8C3)

                            =   2 / 10  =  1 / 5

Hence required conditional probability  =  1 / 5

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