i.
Experienced operator of a company
produced a particular defective item.
ii.
Operator with some experience of a
company produced a particular defective item.
iii.
New operator of a company produced a
particular defective item.
let’s create some events to work with.
D: “operator produces
a defective item”
E: “experienced
operator”
S: “operator with some
experience”
N: “new operator”
With these definitions, the
information in the problem statement can be written as
P(E) = 0.65
P(D/E) = 0.015
P(S) = 0.25
P(D/S) = 0.025
P(N) = 0.1
P(D/N) = 0.05
The probability that an experience operator
produces a defective item, we are interested in the event E and D.
These probabilities lie along the top branch so
P(E and D) = P(D/E) P(E) = (0.015)(0.65)
= 0.0097
=0.97% Answer
To find the probability that a
particular defective item was produced by a Experienced operator, we need to
compute P(E | D).
The appropriate form of Baye’s Theorem is
=
= 0.4642
= 46.42% Answer
To find the probability that a
particular defective item was produced by a Some Experienced operator, we need
to compute P(S | D).
The appropriate form of Baye’s Theorem is
=
= 0.2976
= 29.76% Answer
To find the probability that a
particular defective item was produced by aNew operator, we need to
compute P(N | D).
The appropriate form of Baye’s Theorem is
P
=
= 0.2381
= 23.81% Answer
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