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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.

Probability Theory1

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) =

Probability Theory

=_      _____.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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