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This problem appears very simple, but I am almost positive that it should not be so simple. For the question below I reason that the probability for part a is 0.6 and part b is 0.6 as well. What am I missing?

Q:Hirsbrunner produces tubas and ships them in lots of twenty. Suppose that 60% of all such lots contain no defective tubas, 30% contain one defective, and 10% contain two defectives. Now suppose that a lot is inspected, with two tubas being selected from it at random, and neither is found to be defective.

a) What is the probability that there are no defectives in that lot?

b) Suppose that the inspected lot is from a shipping container that contains 10 lots, and the other 9 lots were not inspected. What is the probability that there are no defectives in that container?

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You need to find the probability that two randomly selected tubas are both nondefective. This will require use of the law of total probability. Then, you will need to use Bayes' formula to determine the conditional probability that the lot is one of the $60\%$ nondefective lots given that both selected tubas are fine. This is usually not the same as the probability of neither selected tuba being defective given that the lot is one of the $60\%$ lots with no defective tubas. –  Dilip Sarwate Sep 26 '12 at 20:44

3 Answers 3

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(a) Let $G$ be the event that the whole lot is good (zero defectives), and let $P$ be the event there were no defectives in the sample of $2$. We are asked for the conditional probability $\Pr(G|P)$, the probability that the whole lot is good given the information that the sample of $2$ had no defectives. By a formula which is likely familiar to you, essentially the definition of conditional probability, we have $$\Pr(G|P)=\frac{\Pr(G\cap P}{\Pr(P)}.$$ It remains to find the probabilities on the right.

We go first for the harder one, $\Pr(P)$. The event $P$ can happen in three ways: (i) the chosen lot has no defectives, and (of course) no defectives are found; (ii) the chosen lot has one defective, and no defectives are found; or (iii) the chosen lot has two defectives, and no defectives are found.

The probability of (i) is clearly $0.6$. Note that this is also $\Pr(P\cap G)$.

The probability that the chosen lot has $1$ defective is $0.3$. If we sample from this lot, then with probability $\frac{19}{20}$ the first tested item is OK, and given that happened, the probability the second item tested is OK is $\frac{18}{19}$, for a product of $\frac{18}{20}$. So the probability of (ii) is $(0.3)(18/20)$, which is exactly $0.27$.

In the same way, we find that the probability of (iii) is $(0.1)(18/20)(17/19)$, which is approximately $0.0805$.

So $\Pr(P)\approx 0.6+0.27+0.0805$, which is about $0.9505$.

Finally, for our conditional probability, divide $\Pr(G\cap P)$ by $\Pr(P)$. We get about $0.63123$.

(b) We are expected to assume independence. For the inspected lot, the probability it has no defectives is approximately $0.63213$. The $9$ uninspected lots each have probability $0.6$ of having no defectives. Multiply.

Remark: Our $0.63123$ is a little bigger than your conjectured answer of $0.6$. This is because the fact that no bads were found in our inadequate inspection makes it somewhat more likely that the lot is an all-good lot.

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For a, you are missing that not finding a defective improves the odds that there is none. The chance of finding no defectives is $60\% (\text{no defectives}) + 30\%\cdot \frac {19\choose 2}{20\choose 2}(\text{one defective and missed it}) \\ +10\%\cdot \frac {18\choose 2}{20\choose 2}(\text{two defective and missed them})\approx 60\%+27\%+8.05\%=95.05\%$ The chance that there are no defectives given the inspection is then $\frac {60}{95.05}\approx 63.12\%$

For b, if you don't inspect any others, the chance that the other nine lots have no defectives is $0.6^9$ and you need to multiply the answer from a by this.

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For (a), imagine that you have ten lots, six with no defectives, three with one defective each, and one with two defectives. In other words, imagine that you’re facing a concrete situation in which the probabilities work out exactly. This is always legitimate in such problems.

If you pick one of the first six to test, there are $\binom{20}2$ ways to pick two tubas to test, and in every case both will be good. If you pick one of the next three lots to test, only $\binom{19}2$ of the $\binom{20}2$ possible samples will contain no defective. And if you choose the last lot, only $\binom{18}2$ of the $\binom{20}2$ possible samples will contain no defective.

Since you actually found no defective tuba in the sample, it must have been one of the $$6\binom{20}2+3\binom{19}2+\binom{18}2$$ possible samples containing no defective tuba. $6\binom{20}2$ of those samples came from lots with no defective tuba, so the probability that the sample came from such a lot is

$$6\frac{\binom{20}2}{6\binom{20}2+3\binom{19}2+\binom{18}2}=\frac{1140}{1806}\approx 0.63123\;.$$

The textbook solution of such problems uses Bayes’ theorem. If $A_0,A_1$, and $A_2$ are the events that the lot contains $0,1$, or $2$ defectives respectively, and $B$ is the event that we find no defective tubas, then Bayes’ theorem says that

$$\begin{align*} \mathrm{Pr}(A_0|B)&=\frac{\mathrm{Pr}(B|A_0)\mathrm{Pr}(A_0)}{\mathrm{Pr}(B|A_0)\mathrm{Pr}(A_0)+\mathrm{Pr}(B|A_1)\mathrm{Pr}(A_1)+\mathrm{Pr}(B|A_2)\mathrm{Pr}(A_2)}\\\\ &=\frac{1\cdot0.6}{1\cdot0.6+\frac{\binom{19}2}{\binom{20}2}\cdot0.3+\frac{\binom{18}2}{\binom{20}2}\cdot0.1}\\\\ &=\frac{0.6}{0.6+\frac{19\cdot18}{20\cdot19}\cdot0.3+\frac{18\cdot17}{20\cdot19}\cdot0.1}\\\\ &\approx0.63123\;. \end{align*}$$

In (b), the answer to (a) gives the probability that the inspected lot contains no defectives. Each of the other nine lots independently has probability $0.6$ of having no defectives, so the probability that all nine have no defectives is only $0.6^9$. To get the probability that there are no defective tubas in the entire container. you have to multiply this figure by the answer to (a); the result is only about $0.00636$, so this inspection proceduce isn’t very effective!

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