Asked by Golu
An urn contains 10 Coupons marked 1,2,3,4,5,6,7,8,9,10 are drawn. Then find the chance that the difference of the coupons exceeds 3 ?
Answers
Answered by
Reiny
I will assume the difference to be between pairs of coupons
possible cases:
5,1
6,1 6,2
7,1 7,2 7,3
8,1 8,2 8,3 8,4
9,1 9,2 9,3 9,4 9,5
10,1 10,2 10,3 10,4 10,5 10,6
total 21
number of pairs = C(10,2) = 45
prob(of your event) = 21/45 = 7/15
possible cases:
5,1
6,1 6,2
7,1 7,2 7,3
8,1 8,2 8,3 8,4
9,1 9,2 9,3 9,4 9,5
10,1 10,2 10,3 10,4 10,5 10,6
total 21
number of pairs = C(10,2) = 45
prob(of your event) = 21/45 = 7/15
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