A group of 3 female student and 5 male students are available to fill
certain student government posts. If 4 students are to be randomly
selected from this group, find the probability that exactly 2 female
students will be chosen among the 4 chosen.

Hello roadapples,
There exactly are 8C4 = 8!/(4!*4!) = 70 ways to select 4 students out
of the group of 8 students.
To solve this problem, we want to know out of those ways, how many of
them contain 2 males and 2 females.
There are 3C2 = 3!/(2! * 1!) = 3 ways to choose 2 female students from
a group of 3.
Likewise, there are 5C2 = 5!(2! * 3!) = 10 ways to choose 2 male
students from a group of 5.
Thus, for each of the 3 ways to choose 2 female students, there are 10
ways to choose the remaining 2 male students, making a total of 3 * 5
= 15 ways to choose 2 female students.
Thus, the probablity that exactly 2 female students are chosen among 4
is 15/70 = 0.2143

I hope that answered your question. If anything is unclear, please
request for a clarification before rating this answer.
secret901-ga

References:
http://www.scit.wlv.ac.uk/university/scit/maths/calculus/modules/topics/precalc/probabil/learn.htm
http://www.stanford.edu/~entzu/Stat%20116/lecture2.pdf problem 1.4b
http://www.montgomerycollege.edu/Departments/mathrv/ma110review.pdf
problem 37

Search Strategy:
I know this from the top of my head.  I located some sample problems
with solutions online to make sure that my method is correct.

Search terms:
committee answer chosen probability
"A group of" "consists of" "randomly selected" "find the probability"

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Clarification of Answer by secret901-ga on 09 Feb 2003 21:54 PST

My apologies, the equation:
5C2 = 5!(2! * 3!)
should read:
5C2 = 5!/(2! * 3!)

Regards,
secret901-ga

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Clarification of Answer by secret901-ga on 09 Feb 2003 22:40 PST

Oops, thank you alan_cam,
There are actually 3 * 10 = 30 ways to choose 2 female students,
making the probability of choosing 2 female students to be 30/70 =
.4285
I don't know where the 5 came from, please excuse the careless error.

Thanks for all the help from everyone!

Excuse me, but I think there is an error here.

For each of the 3 ways to choose 2 female students, there are 10
ways to choose the remaining 2 male students, making a total of 3 * 10
= 30 ways to choose 2 female students.
  
Thus, the probablity that exactly 2 female students are chosen among 4
is 30/70 = 0.4286

The previous comment is correct. The probabilities are as follows:

0 women  = [3C0 * 5C4 / 8C4] =  [1 *  5 / 70] = 0.0714
1 woman  = [3C1 * 5C3 / 8C4] =  [3 * 10 / 70] = 0.4286
2 women  = [3C2 * 5C2 / 8C4] =  [3 * 10 / 70] = 0.4286
3 women  = [3C3 * 5C1 / 8C4] =  [1 *  5 / 70] = 0.0714
Total    =                      [    70 / 70] = 1.0000

When in doubt, make sure all distinct probabilities sum to 1.