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anzhelika [568]
3 years ago
15

A u.s. customs inspector decides to inspect 3 out of 16 shipments for contraband. find the probability that 2 out of the 3 shipm

ents will contain contraband if the selection is random and unknown to the inspector, 5 of the 16 shipments contain contraband.
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0
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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Answer:

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Step-by-step explanation:

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3 0
3 years ago
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posledela
There are two methods to solve this problem. One of them is the typical method and the other one is the short-cut. I will explain the shortcut

Short-cut: 
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\frac{AD}{CE}= \frac{DB}{EB}

Now, you just need to identify the terms, substitute, and solve for the unknown, which would be k. 

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5 0
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