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Problem description

A package contains six batteries. Two of them are defective. Suppose that we randomly select three batteries and test them for defects.

What is the probability that exactly one of the three selected batteries are defective?

Determining the sample space

$ _{n}C_{r} = \frac{n!}{r!(n-r)!} = \; _{6}C_{3} = \frac{6!}{3!(6-3)!} = 20 = S $

The above determines the sample space for the first part of the problem description (we select three batteries).

Question

From manually enumerating the possible combinations from the sample space, like $ S = \{(D1, D2),\ (D1, D3)\ …\ (A5, A6)\} $; I know that the answer to the problem is:

$ P(E) = \frac{s_{e}}{S} = P(12/20) = P(3/5) = P(60 \ \%) $.

But what is the formulaic way to determine that there are $s_{e} = 12$ combinations that can contribute to our event $E$ where exactly one of the two sampled batteries are defective?

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  • $\begingroup$ Your language appears to confound the counts of certain things with the "sample space." Could you clear this up by editing your post to explain what you mean by "sample space"? $\endgroup$
    – whuber ♦
    Commented Apr 18, 2016 at 17:16
  • $\begingroup$ @Antoni There are several effective ways to model this problem--which is one reason I was asking for clarification. The calculations in the question suggest Winterflags is thinking of the sample space as being the set of all three-element samples from the package. However, the question itself refers to "two selected batteries," which contradicts that. Thus we cannot be sure what is being asked. If we assume "two" was intended to be "three," then the event in question consists of all three-element samples containing exactly one defective battery. $\endgroup$
    – whuber ♦
    Commented Apr 20, 2016 at 21:42
  • $\begingroup$ @whuber It was supposed to say three selected batteries. It was a typing mistake that made it two in the text. I'm still not entirely sure what you were trying to make me aware of, but if it was a very winded way of alerting me that there may be a typo in the question, then yes, you were right. Although I don't blame anyone for being confused by the typo since there were two defective batteries. I have fixed it now. $\endgroup$
    – P A N
    Commented Apr 20, 2016 at 22:09
  • $\begingroup$ It's not just a typo (that was the last problem I detected). Many people, for instance, would model this in terms of sequences of three distinct batteries taken from the package. That changes your assertion about the sample space--although of course it does not change the answer to the question. The more significant problem is that you assert the sample space is a number, which it decidedly is not. Finally, your title is a complete mystery. How is it connected with the question? $\endgroup$
    – whuber ♦
    Commented Apr 20, 2016 at 22:14
  • $\begingroup$ @whuber How would you phrase the title? $\endgroup$
    – P A N
    Commented Apr 20, 2016 at 22:17

2 Answers 2

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It seems as though the successful events (samples of 3 without replacement containing exactly 1 defective battery) can be formalized by simply walking through the choices to build up these sets. If we generalize the overall number of batteries to choose from to $N=6$; the sample size to $n=3$; the total number of defective batteries to $B =2$; and the number of defective batteries in the sample as $b=1$,

  1. We look at the number of ways of selecting $b$ defective batteries when there are $B$ to choose from: $B\choose b$, or $2\choose1$ in the OP example.
  2. Then we turn to the number of ways of selecting the rest of the sample constituents, $n-b$, or $3-1$, from among the non-defective batteries, $N-B$, or $6 - 2$. In other words: ${N-B}\choose{n-b}$, or $4\choose2$ choices.

Together, we can formulate the number of successful events as:

$${{B}\choose{b}} {{N-B}\choose{n-b}}$$

As @whuber indicates, we are just working out the numerator in the PMF of the hypergeometric distribution with the random variable $X= b$ (in this case, $b=1$):

$$P(X=b)=\frac{{{B}\choose{b}} {{N-B}\choose{n-b}}}{N\choose n}$$

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There are 2 ways to get one of the bad batteries in the selected group. (2 choose 1)

There are 6 ways (order not important) to distribute the 4 good batteries amongst the selected and non-selected groups. (4 choose 2)

2 x 6 = 12

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