Simple Probability Question

Hi,

I'm a stumped on this (probably easy for most) probability question. I want to know how to solve it, as well as any potential ones after it, but I'm not exactly sure how to begin.

A six person jury will be selected from a pool of nine men and nine women. Supposing the jury is selected at random, what are the odds that all of the jury members will be women?

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My guess, which is probably wrong, is:

9/18 x 8/17 x 7/16 x 6/15 x 5/14 x 4/13 = 0.0045 = 1 in 221 chance all six will be women?

Any help and/or tips to get me started would be appreciated.

Thanks!

Your calculation is correct.

luejinX said:

9/18 x 8/17 x 7/16 x 6/15 x 5/14 x 4/13

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Multiplying out this kind of stuff can be simplified or quickened this way:
a = 4, b = 9 (start and end of numerator)
c = 13, d = 18 (start and end of denominator)

b! * (c - 1)! / [d! * (a - 1)!]

9! * 12! / (18! * 3!) = .0045...

Hello, luejinX!

Your reasoning and your answer are both correct . . . Good work!

Here's another approach.

A 6-person jury will be selected from a pool of 9 men and 9 women.
If the jury is selected at random, what is the probability that all of the jury members will be women?

Click to expand...

There are 18 people from which the 6 jurors will be chosen.
. . \(\displaystyle \text{There are: }\:{18\choose 6} \:=\:\frac{18!}{6!\,12!} \:=\:18,\!564\text{ possible juries.}\)

\(\displaystyle \text{To select 6 women from the available 9 women,}\)
. . \(\displaystyle \text{there are:}\:{9\choose6} \:=\frac{9!}{6!\,3!} \:=\:84\text{ ways.}\)

\(\displaystyle \text{Therefore: }\(\text{6 women}) \;=\;\frac{84}{18,\!564} \;=\;\frac{1}{221}\)

Thanks guys,

I really appreciate the comments.