There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one produced at factory B is defective with probability .01, and each one producedat factory C is defective with probability .05. Suppose you purchase two radios that were produced atthe same factory, which is equally likely to have been any factory. Given the first radio that you checkis defective, find the probability that the other one is also defective.

Answer:The probability is 0.02667Step-by-step explanation:Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:P(D2/D1) = P(D2∩D1)/P(D1)Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:[tex]P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02[/tex]At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:[tex]P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01[/tex]Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:[tex]P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05[/tex]So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:P(A)=1/3P(B)=1/3P(C)=1/3Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:[tex]P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C[/tex]P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)P(D2/D1) = 0.02667