Q.

Three natural numbers are taken at random from the set  A={x/1x100,xN}  and the probability that the A.M of the numbers taken is 25 is  K 100C3  then the digit in units place of k is

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Detailed Solution

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n(s)=100C3 Since AM of 3 numbes is = 25  Then their sum is 75 x1+x2+x3=75,x1x21,x31 Number of +ve integral solutions  74C2=2701 Required probability =  2701 100C3

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Three natural numbers are taken at random from the set  A={x/1≤x≤100,x∈N}  and the probability that the A.M of the numbers taken is 25 is  K 100C3  then the digit in units place of k is