# Finding Conditional Probabilities Problem

There are $3$ production lines in a factory. $2$ of them are new lines and $1$ of them is an old line. All production lines produce at the same rate. The old production line has $8\%$ defective rate while the new production lines have only $3\%$ defective rate. The components are shipped to customers in $100$-unit lots. A buyer received a lot and tested $5$ components and $1$ of it failed.

I want to find the probability that this lot that the buyer had bought was produced by the old line and also the probability that this lot was produced by one of the new lines.

I let the production lines be $A$, $B$ and $C$. Line $C$ is the old one. I also let $D$ be the defective rate. So, I believe the info are given as $P(D|C)=0.08$, $P(D|A)=0.03$ and $P(D|B)=0.03$.

But I am stuck right after this. I don't understand how I should use the $100$ units and $1$ out of $5% tested units in a lot failed. How should I carry on to find out the probability that the lot that the buyer had bought was produced by the old line and also the probability that the lot that the buyer had bought was produced by one of the new lines? - Your defective rates are off, they should be$0.08$,$0.03$. – André Nicolas Apr 23 '12 at 16:00 oh yea...Thanks for alerting me about the typo. I have edited my question. :) – xenon Apr 23 '12 at 16:01 ## 2 Answers Let us find the probability that the lot was produced by the new line, given that exactly$1$of the items failed. So let$X$be the event "exactly$1$item of the$5$tested failed" and let$C$be the event that the lot was produced by the old line. We want$P(C|X)$. By the usual formula for conditional probabilities, we have $$P(C|X)P(X)=P(C\cap X).$$ If we can find$P(X)$and$P(C\cap X)$, we will be finished. The event$X$can happen in$3$different ways: (i) the lot was produced by$A$and$1$item of the$5$tested failed; (ii) the lot was produced by$B$and$1$item failed; (iii) the lot was produced by$C$and$1$item failed. Given that the lot was produced by$A$, what is the probability that$1$item out of$5$tested failed? This is a Binomial situation. The probability of$1$failure in$5$trials is$\binom{5}{1}(0.03)^1(0.97)^4$. Call this number$a$. Given that the lot was produced by$B$, the probability of$1$failure in$5$trials is$\binom{5}{1}(0.03)^1(0.97)^4$, same of course as for$A$. But call this number$b$. Given that the lot was produced by$C$, the probability of$1$failure in$5$trials is$\binom{5}{1}(0.08)^1(0.92)^4$. Call this number$c$. Thus $$P(X)=a\frac{1}{3}+b\frac{1}{3}+c\frac{1}{3}=\frac{a+b+c}{3}.$$ The probability$P(C\cap X)$is somewhat easier to find. I will leave it to you. Remark: Note that the lot size was irrelevant. We assumed independence. This is not fully realistic. There are situations where the overall failure rate is, say,$3\%$, but "out of spec" happens in bursts, a grain of dust that affects several items produced at more or less the same time, or a quality checker taking a little smoke break. - This is an exercise in computing with Bayes' rule. Let$O$represent "produced using the old line" and$N$be "produced using the new line", and let observation$A$be "1 in 5 were faulty." Then you want to compute $$P(O|A) = \frac{P(O)P(A|O)}{P(A)}$$ $$P(N|A) = \frac{P(N)P(A|N)}{P(A)}$$ Notice that these two expressions partition the event space, so they must add up to one, and also that the denominators are identical. So we can ignore the denominator when doing the calculation, which is a nice optimisation. You already know that$P(O)=1/3$and$P(N)=2/3$. Can you work out how to compute$P(A|N)$and$P(A|O)\$, given the defect rates?

Hint: Binomial random variables.

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