Customs agents suspect that 3 containers on a recently arrived ship may contain

Question

Customs agents suspect that 3 containers on a recently arrived ship may contain food products that must be quarantined. The 1st container holds 8 boxed, two of which contain food products that must be quarantined. The 2nd container holds 10 boxes, 2 of which contain food products that must be quarantined. And the 3rd container holds three boxes, 1 of which contains food products that must be quarantined.

a) If 5 boxes are selected at random, what is the probability taht at least 2 will contain food products that must be quarantined?

b) Now suppose that the container is selected at random, and then a box is selected at random from the container. If the selected box does not contain food products that must be quarantined, what is the probability that the box was selected from the second container?

Explanation / Answer

There are 3 containers
Container 1: 8 boxes = 6 + 2 quarantined
Container 2: 10 boxes = 8 + 2 quarantined
Container 3: 3 boxes = 2 + 1 quarantined

(A)
there are total 21 boxes, out of which 5 boxes can be selected in 21C5 ways
There are total 5 quarantined boxes.

Let A be the event that less than 2 quarantined boxes are selected.
Hence P(A) = 5C0*16C5/21C5 + 5C1*16C4/21C5 = 0.2147 + 0.4472 = 0.6619

(B)
Probability that a non-quarantined box is selected from container 1, P(Q1'|C1) = 6/8
Probability that a non-quarantined box is selected from container 2, P(Q2'|C2) = 8/10
Probability that a non-quarantined box is selected from container 3, P(Q3'|C3) = 1/3

Hence required probability that if the probability that the box was selected from the second container

P(C2|Q2') = P(Q2'|C2) / (P(Q1'|C1) + P(Q2'|C2) + P(Q3'|C3)) = (8/10)/(6/8 + 8/10 + 1/3) = 0.4248

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