Question

Therearetwolocalfactoriesthatproduceradios.EachradioproducedatfactoryA is defective with probability .05, whereas each one produced at factory B is defective with probability .01. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the first radio that you check is defective, what is the conditional probability that the other one is also defective?

Answer 1

Answer: 0.043

Explanation:

A="The radio is from factory A"

B="The radio is from factory B"

Di="The i-th radio is defective"

P(D|A)=0.05

P(D|B)=0.01

P(A)=1/2

P(D∩A)=P(D|A)P(A)=0.05*1/2=0.025

P(D∩B)=P(D|B)P(B)=0.01*1/2=5x10^-3

P(D1∩D2|A)=0.05*0.05=2.5x10^-3

P(D1∩D2'|A)=0.05*0.95=0.0475

P(D1'∩D2|A)=0.05*0.95=0.0475

P(D1'∩D2'|A)=0.95*0.95=0.9025

P(D1∩D2|B)=0.01*0.01=1x10^-4

P(D1∩D2'|B)=0.01*0.99=9.9x10^-3

P(D1'∩D2|B)=0.01*0.99=9.9x10^-3

P(D1'∩D2'|B)=0.99*0.99=0.9801

P(D1∩D2∩A)=0.05*0.05*1/2=1.25x10^-3

P(D1∩D2∩B)=0.01*0.01*1/2=5x10^-5

P(D2|D1)=P(D2∩D1)/P(D1)

P(D1)=P(D1|A)+P(D1|B)=0.05+0.01=0.06

P(D2∩D1)=P(D2∩D1|A)+P(D2∩D1|B)=0.05*0.05+0.01+0.01= 2.6x10-3

P(D2|D1)=2.6x10-3/0.06=0.043