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I have this statistics problem where it's a Venn diagram relation. There's

88 people in total

21 belong to category A

17 belong to category B

11 belong to both A and B

39 belong to no category

I need to calculate the probability that when 3 people are selected at random from the whole sample, none belongs to category A.

When I do this, do I do

$88-(21+11)=56$

$(56/88)(55/87)(54/86)=(315/1247)$?

Do I need to subtract 1 from the denominator each time like I did to ensure the same person doesn't get chosen twice?

Is there another way to do this? This method seems odd to me because what if the question asks "30 people are chosen at random, what's the probability that no one is in group A"? Wouldn't that, then, take me ages to multiply all the probabilities together?

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  • $\begingroup$ It would take a long time to multiply... if you choose to write each and every entry in the multiplication out manually. However we have ways of writing long and tedious multiplications in very compact forms. $n!=1\cdot 2\cdot 3\cdot 4\cdots (n-1)\cdot n$, so if you want to multiply $k\cdot (k+1)\cdot (k+2)\cdots (n-1)\cdot n$ you can write it as $\frac{n!}{(k-1)!}$. It may be easier as well to treat it as though the order of selection doesn't matter, and instead use binomial coefficients. $\endgroup$
    – JMoravitz
    Commented Jun 19, 2016 at 3:25
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    $\begingroup$ It should be mentioned that the data you provide makes no sense if we interpret the phrase "21 belong to category A" as belonging to category A and possibly others too. The numbers only work in this case if it is "21 belong to category A and only category A" With that in mind, your calculation is correct. It could have been written more compactly as $\frac{\binom{56}{3}}{\binom{88}{3}}$. For thirty people, $\frac{\binom{56}{30}}{\binom{88}{30}}$ $\endgroup$
    – JMoravitz
    Commented Jun 19, 2016 at 3:29

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Are your sure it's 88 people altogether, not 66?

I get four disjoint categories $A \cap B = AB$ with 11, $BA^c$ with 6, $AB^c$ with 10, and $(A\cup B)^c$ with 39.

Then 11 + 6 + 10 + 39 = 66.

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  • $\begingroup$ Presumably the “21 in category A” and “17 in category B” mean “21 in category A only” and “17 in category B only.” Then you get 88 total people. $\endgroup$
    – Steve Kass
    Commented Jun 22, 2016 at 1:20
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    $\begingroup$ OK so "A" really means $AB^c$ in the first part. What do you suppose it means in the 2nd part where you're choosing to avoid "A"? Also I guess you're choosing without replacement. Maybe context of what topics came just before this problem will help with some of that. Even so, this has to be one of the worst-written problems I've seen. Hope the rest in your course are clearer. $\endgroup$
    – BruceET
    Commented Jun 22, 2016 at 4:26

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