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I am currently doing a self-study on Conditional Probability and faced the below question.

A manufacturing plant makes radios that each contains an integrated circuit (IC) supplied by three sources A, B and C. The probability that the IC in a radio came from one of the sources is 1/3, the same is true for all sources. ICs are known to be defective with probabilities 0.001, 0.003 and 0.002 for sources A, B and C respectively.

What is the probability any given radio will contain a defective IC?

Approach-

Say, Rd - event of radio being defective

Given P(Rd|A) = 0.001, P(Rd|B) = 0.003 and P(Rd|C) = 0.002;

P(A) = P(B) = P(C) = 1/3

We have to find P(Rd) = probability of finding a defective IC in the radio

= P(Rd|A)*P(A) + P(Rd|B)*P(B) + P(Rd|C)*P(C)

P(Rd) = 1/3*(0.001 + 0.003 + 0.002) = 0.002

But the answer given is 0.005978. Not sure what's wrong with the approach I have taken. Any help is highly appreciated.

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    $\begingroup$ I concur with your approach. Have you carefully checked, the question is exactlyI the one you gave us? Is the source of the answer reliable? $\endgroup$
    – Bernhard
    Commented Oct 5, 2017 at 11:24
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    $\begingroup$ @Xi'an, it's best not to add the [ss] tag for the OP, but to ask them to add it themselves & read its wiki (there is suggested text here). This way they are more likely to be familiar w/ our policies & have demonstrated some buy-in that they will abide by them. If they don't add the tag after being prompted, we can close the thread. $\endgroup$
    – gung - Reinstate Monica
    Commented Oct 5, 2017 at 19:17
  • $\begingroup$ I agree with Bernhard, double check the question and answer and let us know what you found. $\endgroup$
    – Samuel
    Commented Oct 5, 2017 at 21:04
  • $\begingroup$ This question is from Problems and Solutions in Probability, Random Variables and Random Signal Principles by Peyton Peebles. I have copied the question verbatim. $\endgroup$
    – SrikanthRaja
    Commented Oct 6, 2017 at 2:29

1 Answer 1

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Here's the solution given in the book.

Given P(A) = P(B) = P(C) = 1/3 (since equi-probable)

P(A is defective) = 0.001 then P(A is not defective) = 0.999

P(B is defective) = 0.003 then P(B is not defective) = 0.997

P(C is defective) = 0.002 then P(C is not defective) = 0.998

Probability that any given radio is defective IC

= P(A defective) * P(B and C are not defective) + P(B defective) * P(A and B are not defective) + P(C defective)*P(A and B are not defective)

A, B and C are independent so P(ABC) = P(A)P(B)P(C)

= P(A defective)*P(B not defective)*P(C not defective) + P(B defective)*P(C not defective)*P(A not defective) + P(C defective)*P(B not defective)*P(A not defective)

= 0.001*0.997*0.998 + 0.003*0.998*0.999 + 0.002*0.999*0.997

= 0.005978

There is a part b to the question

b) If a radio contains a defective IC, find the probability that it comes from source A.

Answer

P(A|defective) = P(A and defective)/P(defective) = P(A)*P(defective|A)/P(defective)

= (1/3)*0.001/(0.001+0.003+0.002) = 0.0555

Looks like a radio can contain more than one IC. Is the question asking What is the probability any given radio will contain ONE defective IC for part a of the question? I'm also finding it hard to undertand how P(defective) = P(A is defective) + P(B is defective) + P(C is defective) in the denominator to answer b

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  • $\begingroup$ This answer, even though it has been provided by the textbook author, is grossly incorrect. $\endgroup$
    – Dilip Sarwate
    Commented Nov 22, 2020 at 1:18
  • $\begingroup$ @Dilip Sarwate, thanks for the confirmation $\endgroup$
    – SrikanthRaja
    Commented Nov 23, 2020 at 16:01

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