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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:

1: first radio is defective 2:second radio is defective

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?

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

1 Answer 1

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The second solution considers that the "first radio you checked is defective" means "at least one is defective."   However, it doesn't.

What it actually means is "at least one is defective and it is that specific one — the one we tested."

(tl;dr)   The probability of that happening is what you calculated.   You were correct.

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