# Conditional Probability Problem: Two Radios from Two Factories

Q: There are two local factories that produce radios. Each radio produced at factory A 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?

My solution:

P(2|1)=P(1∩2)/P(1)

P(1∩2)=P(both are defective)=0.05*0.05*0.5+0.01*0.01*0.5=13/10000

P(1)=0.5*0.05+0.5*0.01=0.03

P(2|1)=(13/10000)/0.03=0.0433

However I found another solution online:

## from A:

P[both defective] = .05*.05 = 0.25% P[one defective] = 2C1*.05*.95 = 9.5% P[at least one defective] = sum of the above = 9.75% ............ this is the sample space

so P[both defective | one defective] = 0.25/9.75 = 2.564%

## from B:

P[both defective] = .01*.01 = 0.01% P[one defective] = 2C1*.01*.99 = 1.98% P[at least one defective] = sum of the above = 1.99% ............ this is the sample space

so P[both defective | one defective] = 0.01/1.99 = .5025%

## combined result

there is equal probability of choosing factory A or B, so P[both defective | one defective] = 0.5(1.99 + .5025) %

## = 1.246%

Can anyone tell me which solution is correct?

• I think yours is right except you have a typo (twice): $13/1000$ should be $13/10000$. Feb 9, 2016 at 20:47