Permutations & Combinations: Bankteller Problem

rickdundee
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Homework Statement


There are 6 males and 4 females awaiting to see a teller at a bank.

Only 4 people can be served at one time.
1) How many ways can four of the people be picked and served one at a time, if they must include two(2) men and two(2) women?


2) If indeed the four people are picked randomly, what is the probability that the four will include two (2) men and two (2) women?
This is the question I am confused about.


The Attempt at a Solution


My solution for Problem#1: (6 choose 2) * (4 choose 2) = 90
 
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How many ways can you pick 4 people out of the 4 men and 6 women (with no restriction on how many men or women there are)?
 
That would be 10c4
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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