Probability Theory Sample Assignment
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Question: A shopkeeper’s buys a particular kind of lights bulb from these manufacturers A1, A2 and A3. He buys 30% of his stock from A1, 45% from A2 and 25% from A3. In the past he has found that 2% of A3’s bulb are faulty whereas only 1% of A1’s is and A2’s are. Suppose that he chooses a bulb and finds it is faulty. what is the probability that it was one of A3’s bulb?
Sol: Here P(A1), P(A2) and P(A3) are prior probability because they exist before we gain any information from the experiment itself. If he picks a bulb at random.
P (A1) = 0.3 P (A2) = 0.45 & P (A3) = 0.25
The probability of faulty bulb will be grater than P (A3) = 0.25. Since A3 produces a greater proportion of faulty bulbs than A1 & A2. If F is the event that the bulb is faulty, then P (A3/F) is the probability that we require. We have
P (F/A1) = .01, P (F/A2) = .01 P (F/A3) = .02
This information is shown on a tree diagram.
P (A3/F) = P (A3) P (F/A3) ____________
P (A1) P (F/A1) + P (A2) P (F/A2) + P (A3) P (F/A3)
by using Barye’s The. P (B1/A) =
=_ _____.25 x .02 ____ = 0.4
.3 x .01 + .45 x .01 + .25 x .02
Similarly, if we wish to know the probability that the faulty bulb was supplied by A1 or A2, we have
P (A1/F) =__ P(A1) P(F/A1) _____________
P (A1) x P (F/A1) + P(A2) x P (F/A2) + P (A3) x P (F/A3)
=_______ .3 x .01__ ____ = 0.24
.3 X .01 + .45 X .01 + .25 x .02
and since A1, A2 and A3 are mutually exclusive
P(A2/F) = 1-0.4-0.240 = .36
Probability Theory Sample Assignment
Question: A shopkeeper’s buys a particular kind of lights bulb from these manufacturers A1, A2 and A3. He buys 30% of his stock from A1, 45% from A2 and 25% from A3. In the past he has found that 2% of A3’s bulb are faulty whereas only 1% of A1’s is and A2’s are. Suppose that he chooses a bulb and finds it is faulty. what is the probability that it was one of A3’s bulb?
Sol: Here P(A1), P(A2) and P(A3) are prior probability because they exist before we gain any information from the experiment itself. If he picks a bulb at random.
P (A1) = 0.3 P (A2) = 0.45 & P (A3) = 0.25
The probability of faulty bulb will be grater than P (A3) = 0.25. Since A3 produces a greater proportion of faulty bulbs than A1 & A2. If F is the event that the bulb is faulty, then P (A3/F) is the probability that we require. We have
P (F/A1) = .01, P (F/A2) = .01 P (F/A3) = .02
This information is shown on a tree diagram.
P (A3/F) = P (A3) P (F/A3) ____________
P (A1) P (F/A1) + P (A2) P (F/A2) + P (A3) P (F/A3)
by using Barye’s The. P (B1/A) =
=_ _____.25 x .02 ____ = 0.4
.3 x .01 + .45 x .01 + .25 x .02
Similarly, if we wish to know the probability that the faulty bulb was supplied by A1 or A2, we have
P (A1/F) =__ P(A1) P(F/A1) _____________
P (A1) x P (F/A1) + P(A2) x P (F/A2) + P (A3) x P (F/A3)
=_______ .3 x .01__ ____ = 0.24
.3 X .01 + .45 X .01 + .25 x .02
and since A1, A2 and A3 are mutually exclusive
P(A2/F) = 1-0.4-0.240 = .36
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