I have the following question from a book.

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is?

And the solution to this is:

The problem can be done using binomial distribution since the population is infinite.

Probability of defective item $p=0.1$. Probability of non-defective item $q=0.9$. Probability that exactly 2 of the chosen items are defective:

$\newcommand\Mycomb[2][^{10}]{{#1\mkern-0.5mu}{}C_{#2}}$ $=\Mycomb{2}p^{2}q^{8}=\Mycomb{2}(0.1)^{2}(0.9)^{8}=0.1937$

Now my doubt is how can the probability of the new random section of 10 items have the same as that of the bigger (infinitely in this case, although I would prefer a general answer) set? Suppose we coincidentally get 10 defective items then the probability of the lot of 10 having defective items would be $100 \%$, right?

Can someone please explain to me if the solution is wrong or I am missing something here?

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    $\begingroup$ That doesn't make sense. You need to clarify your assumptions. If you are assuming that each draw is an independent trial with "success" probability $\frac 1{10}$ then this is just a routine binomial distribution. This, as I say, is what the official solution appears to do. If you are assuming something else you need to say what it is. $\endgroup$ – lulu Aug 29 '18 at 18:23
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    $\begingroup$ To stress: at no point is anybody assuming that the smaller collection has the same statistics as the larger. Indeed, if we were to make that assumption the answer would be $0$ since $10\%$ of $10$ is $1$. $\endgroup$ – lulu Aug 29 '18 at 18:24
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    $\begingroup$ Yes, that's a perfect way to look at it (at least, under the assumption the problem appears to be making). Just imagine that Heads means defective, and occurs with probability $\frac 1{10}$. $\endgroup$ – lulu Aug 29 '18 at 18:49
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    $\begingroup$ Just to stress: there is no notion that the statistics of a sample has to follow the statistics of the large (maybe infinite) pool that produced it. Thus if you toss a fair coin $100$ times there is no notion that you will always get exactly $50$ heads. Indeed, that is a rather low probability event (about $0.08$ using the binomial distribution). To be sure, if the sample is "large enough" then one expects the sample statistics to be "close to" the "true" statistics, but that is another matter, certainly not relevant to very small samples (as here). $\endgroup$ – lulu Aug 29 '18 at 18:55
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    $\begingroup$ As a possibly useful exercise, solve the problem assuming a specific lot size. As has been remarked, the binomial distribution method is just an approximation, though one that becomes extremely accurate if the lot size is large. I already pointed out that if the lot size is $10$ the answer is $0$. What if it is $20$? $100$? $1000$? $10^6$? You should see that the answer rapidly converges to the one given by the binomial distribution. And, since we have no information regarding the lot size, we have to assume something. $\endgroup$ – lulu Aug 30 '18 at 11:27

As stated, the problem is ambiguous. If the lot size is small then we can't treat each choice of an item as an independent event. To take an extreme case, suppose the lot only had $10$ items, with $1$ defective one. In that case, of course, the probability of drawing two defective ones is $0$.

If we fix the lot size as $N$, then there are $\binom N{10}$ ways to choose $10$ items and $\binom {.1N}{2}\times \binom {.9N}{8}$ ways to choose them so that exactly $2$ are defective. Under this assumption the probability, as a function of $N$ is the quotient $$\frac {\binom {.1N}{2}\times \binom {.9N}{8}}{\binom N{10}}$$

In particular, for $N=100$ we get $0.201509885$. For $N=10^6$ we get $0.193710998$.

It seems reasonable to imagine that we are approaching a limit as $N\to \infty$. Indeed, if we assume that $N$ is very large then it makes sense to regard each draw as an independent event. Under that assumption we are free to think of this as a binomial distribution, with "success" probability $.1$ in which case the answer would be $$\binom {10}2\times .1^2\times .9^8=0.193710245$$

which, unsurprisingly, is more or less the same as the value we got for $N=10^6$. After all, while it is technically true that drawing a defective item (from the $100000$ defectives) lowers the probability that the next one is also defective, it certianly doesn't lower it by much.

As an analogy, suppose we were tossing a biased coin repeatedly, and writing down the results. Here our coin comes up $H$ with probability $.1$ and, of course, each toss is independent of all the others.

  • $\begingroup$ Great answer. Thank you so much. $\endgroup$ – paulplusx Aug 31 '18 at 13:04

You wrote:

Suppose we coincidentally get 10 defective items then the probability of the lot of 10 having defective items would be 100%, right?

Yes, but what is the probability of "coincidentally" getting ten out of ten defective items? It would be the probability that the first item is defective (which is $0.1$), times the probability that the second item is defective (which is also $0.1$), times the probability that the third item is defective, etc. In the end, you obtain $(0.1)^{10}$ as the probability that the entire sample is defective.

A somewhat more complex question regards the probability of "coincidentally" getting two defective items out of the sample of ten. One way would be to get defective items for the first two (which happens with probability $(0.1)^2$), and then satisfactory items for the remaining eight (which happens with probability $(0.1)^8$), for a total probability of $(0.1)^2(0.9)^8$.

Of course, there are other combinations than just "first two defective, remaining eight satisfactory." How many combinations are there? Well, there are as many as there are ways to choose two items out of ten, which is $\binom{10}{2} = 45$. Each one has the same probability of $(0.1)^2(0.9)^8$, which means the total overall probability is $45(0.1)^2(0.9)^8$.

It should be emphasized that this reasoning only works because the population is infinite. If it were finite, then the probability distribution of the second item in the sample would depend to some extent on whether the first item in the sample was defective or satisfactory.

  • $\begingroup$ My question is when we make 10-item sample, won't we change the probability of the items currently present in that sample? In your second paragraph you assume that the items have the same probability of being defective as if they are in the original lot. If they were coins I could have understood but they are not and they can only be either defective or not every time you pick. $\endgroup$ – paulplusx Aug 30 '18 at 3:45
  • $\begingroup$ Could you please join this chat. $\endgroup$ – paulplusx Aug 30 '18 at 3:58

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