Answer: (a) 0.002 (b) [tex]\dfrac{1}{6}[/tex]
Step-by-step explanation:
Given : The probability that the IC in a radio came from one of the sources =[tex]P(A)=P(B)=P(C)=\dfrac{1}{3}[/tex]
[tex]P(D|A)=0.001,\ \ P(D|B)=0.003,\ \ P(D|C)=0.002[/tex]
By using the law of total probability :-
[tex]P(D)=P(A)\cdot P(D|A)+P(B)\cdot P(D|B)+P(C)\cdot P(D|C)\\\\\Rightarrow\ P(D)=\dfrac{1}{3}\cdot0.001+\dfrac{1}{3}\cdot0.003+\dfrac{1}{3}\cdot0.002\\\\\Rightarrow\ P(D)=0.002[/tex]
Hence, the probability that any given radio will contain a defective IC : 0.002
By using Bayes theorem , we have
[tex]P(A|D)=\dfrac{P(A)\cdot P(D|A)}{P(D)}\\\\\Rightarrow\ P(A|D)=\dfrac{\dfrac{1}{3}\cdot0.001}{0.002}\\\\\Rightarrow\ P(A|D)=\dfrac{1}{6}[/tex]
Hence, If a radio contains a defective IC, find the probability that it came from company A : [tex]\dfrac{1}{6}[/tex]
