Question 1205263
<pre>
As you have stated the problem, the probability is 1, because since there are 7
males, there are of course 6 males, and since there are 5 females there are of
course 3 females.

So you left out the word "particular". Your question should be:

What is the probability that 6 PARTICULAR male and 3 PARTICULAR female students
ARE included in {{{cross(to)}}} the meeting.

So the 20 females consist of 5 particular females, and 15 non-particular females.
And, the 30 males consist of 6 particular males, and 24 non-particular males.

The number of successful choices are 
1. Choose all 3 particular females 1 way and 2 non-particular females 
C(15,2)=105 ways.
2. Choose all 6 particular males 1 way, and 1 non-particular male C(24,1)=24
ways.
That's (105)(24) = 2520 successful ways.

The number of possible choices are 
1. Choose any 5 females C(20,5)=15504 ways.
2. Choose any 7 males C(30,7)=2035800 ways
ways.
That's (15504)(2035800) = 31563043200 possible ways.

So the probability is 2520/31563043200 which reduces to
7/87675120 or about 0.00000008

Edwin</pre>