Self Study - Conditional Probability 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. 
 A: 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
