Simple Probability Question

[luejinX]

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?

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!

 

[royhaas]

Your calculation is correct.

 

[Denis]

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

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...

 

[soroban]

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?

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

 

[luejinX]

Thanks guys,

I really appreciate the comments.