Answer:
Answer in the pic with steps
Hope it helps
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%
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C(n,k) = n!/(k!(n-k)!)
Answer:
Step-by-step explanation:
You are correct. It's 8/5
.21 as a percentage is 21%
You can divide 7.98 by 21 then multiply by 100 which gives you an answer of 38.
Hope this helps :)
