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If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?

So there are 10 men and 5 women total.

We want to know the probability that the jury will have at least 8 men.

So the 1-X route seems best here.

So the indirect method is this:

total number of juries that could be randomly selected is: 15!/(12!3!) = 15 * 14 * 13 / (6) = 455

The only way the jury will have fewer than 8 men is if a jury of 7 men and 5 women (the maximum number of women available) is selected. There cannot be fewer than 7 men on the jury, since the jury must have 12 members and only 5 women are available to serve on the jury.

So what's the prob of there being 7 men on the jury?

10!/(7!3!) = 120

So prob that the jury will be composed of fewer than 8 men is : 120/455 = 24/91

So the prob that there will be at least 8 men is: 1 - 24/91 = 67/91

Is there a direct method to this?

Remember, formula for combinations is: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Jwan622
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1 Answers1

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For the (tedious) direct method, we explicitly consider the juries case by case. There are only three ways for there to be at least $8$ men: $8$ men and $4$ women, $9$ men and $3$ women, and $10$ men and $2$ women. This yields: $$ \frac{\binom{10}{8}\binom{5}{4} + \binom{10}{9}\binom{5}{3} + \binom{10}{10}\binom{5}{2}}{\binom{15}{12}} = \frac{67}{91} $$ where I use the notation: $$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $$

Adriano
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  • Thanks! How do I convert my gross looking combination math to your great-looking binomial notation? – Jwan622 Dec 15 '13 at 04:04
  • Right click on the above equations, go to "Show Math As", then click on "TeX Commands". For example, try cutting-and-pasting the following code into your post: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ – Adriano Dec 15 '13 at 07:05