Faulty Microchip [Probability Problem]

[SEstudent22]

Here is the problem:

" One microchip is being produced in three series, each series produces 18 microchips. In the first series 12 are faulty, in the second series 14 are faulty and in the third series 15 are faulty. A series is picked randomly and one microchip from that series (also randomly). It turns out that the picked microchip is faulty. The same microchip is returned to the same series it was taken from and from that series another microchip is randomly selected. What is the probability that the second microchip is also faulty? "



Here is my work and thinking:

We have three hypotheses and their respected probabilities:
H1 - we pick a chip from the first series
P(H1) = 18/54 = 1/3
H2 - we pick a chip from the second series
P(H2) = 18/54 = 1/3
H3 - we pick a chip from the third series
P(H3) = 18/54 = 1/3

I define the following event:
A - the first picked chip is faulty
I calculate this event's probability using the law of total probability
P(A) = [imath]\sum_{i=1}^{3}P(Hi)P(A|Hi)[/imath] = [imath]\frac{1}{3} * \frac{12}{18} + \frac{1}{3} * \frac{14}{18} + \frac{1}{3} * \frac{15}{18} = 0.76[/imath]


If we randomly take out a chip, the probability that it's faulty is 0.76. Now since that chip is returned to the same series and we try taking a chip again will it once again be 0.76? Then the probability that the second microchip is also faulty would be 0.76^2.
But since when we take out the second chip, we already know which series we're taking a chip from, wouldn't that change the probability and the final outcome?

 

[Dr.Peterson]

If we randomly take out a chip, the probability that it's faulty is 0.76. Now since that chip is returned to the same series and we try taking a chip again will it once again be 0.76? Then the probability that the second microchip is also faulty would be 0.76^2.
But since when we take out the second chip, we already know which series we're taking a chip from, wouldn't that change the probability and the final outcome?

Yes, it would. Also, they're asking for the probability given that the first was faulty, not for the probability that two chips, each chosen by the same routine, are both faulty.

I would try to find the probability that the first chip was taken from each of the series, and then use those new probabilities to find the total probability for the second selection. in place of the 1/3's.

 

[SEstudent22]

Yes, it would. Also, they're asking for the probability given that the first was faulty, not for the probability that two chips, each chosen by the same routine, are both faulty.

I would try to find the probability that the first chip was taken from each of the series, and then use those new probabilities to find the total probability for the second selection. in place of the 1/3's.

I think I understand. So we use Bayes' theorem to find the probabilities that the faulty chip came from each of the series, then we multiply them with their respected probability (to pick a faulty chip) and add them up. Basically "the probability that the chip was taken from the series" times "the probability that a selected chip from that series is faulty". Is this correct?

I was stuck on the question "how would I find out which series the chip was taken from".