Au. s. customs inspector decides to inspect 3 out of 16 shipments for contraband. find the probability that 2 out of the 3 shipments will contain contraband if the selection is random and unknown to the inspector, 5 of the 16 shipments contain contraband.

There are C(5,2) = 10 ways to choose 2 contraband shipments from the 5. There are C(11, 1) = 11 ways to choose a non-contraband shipment from the 11 that are not contraband. Hence there are 10*11 = 110 ways to choose 3 shipments that have 2 contraband shipments among them.

There are C(16,3) = 560 ways to choose 3 shipments from 16. The probability that 2 of those 3 will contain contraband is
  110/560 = 11/56 ≈ 19.6%


C(n,k) = n!/(k!(n-k)!)

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