Asked by Goku
After packing k boxes (numbered 1,2, ..., k) of m items each, workers discovered that
one defective item had slipped in among the km items packed. In an attempt to nd
the defective item, they randomly select n items from each box and examine these.
(a). Find the conditional probability that the defective item is in box 1, given that it
is not found in box 1.
(b). Find the conditional probability that the defective item is in box 2, given that it
is not found in box 1.
one defective item had slipped in among the km items packed. In an attempt to nd
the defective item, they randomly select n items from each box and examine these.
(a). Find the conditional probability that the defective item is in box 1, given that it
is not found in box 1.
(b). Find the conditional probability that the defective item is in box 2, given that it
is not found in box 1.
Answers
Answered by
MathMate
(a) "Given that it is not found in box 1" means that it is not in box 1, so probability of finding it in box 1 is zero.
(b) Conditional probability:
P(A|B)=P(A∩B)/P(B).
Event A: defective in box 2
Event B: defective not in box 1
P(A)=1 box out of k = 1/k
P(B)=not in one of k boxes = (k-1)/k
P(A∩B)=P(A)=1/k
because if it is in box 2, then it implies that it is not in box 1,
i.e. event A is contained in event B.
P(A|B)=P(A∩B)/P(B)
=(1/k)÷((k-1)/k)
=1/(k-1)
Actually this result can be obtained by intuition, because out of k boxes, knowing that item is not in box 1, there are only (k-1) boxes left, hence P(A|B)=1/(k-1)!
(b) Conditional probability:
P(A|B)=P(A∩B)/P(B).
Event A: defective in box 2
Event B: defective not in box 1
P(A)=1 box out of k = 1/k
P(B)=not in one of k boxes = (k-1)/k
P(A∩B)=P(A)=1/k
because if it is in box 2, then it implies that it is not in box 1,
i.e. event A is contained in event B.
P(A|B)=P(A∩B)/P(B)
=(1/k)÷((k-1)/k)
=1/(k-1)
Actually this result can be obtained by intuition, because out of k boxes, knowing that item is not in box 1, there are only (k-1) boxes left, hence P(A|B)=1/(k-1)!
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