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I am studying probabilities and I was intrigued by the following problem:

A laboratory studies daily six subjects from company $A$ and four subjects from company $B$. Between the subjects from company $A$ there are two defective ones and between those of company $B$ there exists one defective subject. One day two subjects out of ten are drawn randomly, to be studied. What is the probability that we choose at least one defective from company $A$, if it is known that we chose a subject from company $A$ during the first draw? (Take into account the case with and without repositioning)

To me it looks like a Bayes probability, but it is quite confusing the way it is presented. I tried to calculate the probability that we choose a defective one from company $A$ during the first draw: $P(A)=P(\text{Choose from company A})P(\text{Subject defective})+P(\text{Choose from company B})P(\text{Subject not defective})$

and do the same for the second draw and eventually use the conditional probability. But it cannot get me anywhere.

Thank you.

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    Not following. First you say we choose six , then you say we have a sample of ten. Where did the ten come from? What rules were followed in selecting them? – 2017-02-02
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    It is not mentioned that we choose six. What it says is the laboratory studies six from $A$ and four from $B$, therefore ten in total. I know, I was confused too in the beginning. – 2017-02-02
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    Got it. And what is meant by the "first draw"? Do you mean that at least one of the two you pick (out of the ten) is from $A$? – 2017-02-02
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    What exactly is repositioning? – 2017-02-02
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    @lulu My guess is that the subjects are not picked simultaneously, that is why two picks are needed. – 2017-02-02
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    @jan I think by repositioning it means taking into account replacing the subject back to the "pool" after studying it or not – 2017-02-02
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    I sympathize...it's hard to work these problems where most of the effort just goes into sorting out what the question means. Are you satisfied with the posted solutions? – 2017-02-02
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    since they actually said 'one day two subjects out of ten are drawn randomly, to be studied. ' I can't see how that can include a case with any replacement, unless 'two' can mean the same subject studied twice, which I guess on reflection is just about possible, although they really needed to explain that better. – 2017-02-02
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    they can't chose simultaneously if the first chosen is from A, which is different from saying 'at least one of them is from A' and again different from 'exactly 1 was from A' - all too often, as people are saying, these points are not clarified with an example by the question setter – 2017-02-02

3 Answers 3

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I am not sure what is repositioning, but what is wrong with the following.

In the first draw, we are drawing from $6$ $A$-company subjects among which $2$ are defective. Hence with $\frac{1}{3}$ we choose defective subject. What we choose subsequently does not matter since we have chosen at least one defective already. With probability $\frac{2}{3}$ we choose non-defective subject and we draw again. In the second draw, we have $9$ subjects with $3$ defective ones. Hence with probability $\frac{1}{3}$ we choose defective.

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    you can choose a defective from A on the second draw, which adds to the 1/3 chance of getting one on the first draw, so it is 1/3 + (2/3)(3/9) =5/9 – 2017-02-02
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P(first drawn from A and a defective from A drawn in either draw) = $\frac{6}{10}(\frac{1}{3} + \frac{2}{3}\times\frac{3}{9}) = \frac{1}{3}$

P(first drawn from A) = $\frac{6}{10} = \frac{3}{5}$

P(a defective from A drawn in either draw|first drawn comes from A) = P(first drawn from A and a defective from A drawn in either draw) / P(first drawn from A) = $\frac{1}{3}/\frac{3}{5} = \frac{5}{9}$

my answer = $\frac{5}{9}$

note this is a case where 2 different are chosen, without replacement

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Assuming "repositioning" is a non standard term for "replacement".

I am also assuming that the question is asking: "Suppose you first choose a unit from $A$ and then you choose a unit from the remaining total (either replacing the first person chosen or not), what is the probability that you get at least one defective unit?" I expect others might interpret the question differently.

Note: in both cases we work from the complement. That is, we compute the probability that we get $0$ defectives.

Case I: Without Replacement

Then to get no defectives we need the first draw (from $A$) to be intact and then we need the second draw (from the total) to be intact. Thus the probability of getting no defectives is $$\frac 46\times \frac 69=\frac 49$$

Thus the answer in this case is $$1-\frac 49=\frac 59$$

Case II: With Replacement

This time the probability that both draws are intact is $$\frac 46 \times \frac 7{10}=\frac 7{15}$$

Thus the answer in this case is $$1-\frac 7{15}=\frac {8}{15}$$