It's discrete. What we're doing is taking a sample of four pulls from a binomial distribution. If we call roosters successes then we're asking what's the probability of one success in four pulls. The formula would be (the number of combinations of one success in four tries) times (the probability of success)^1 times (the probability of failure)^(4-1). Since the prob of success and prob of failure are here both assumed to be 1/2 them the prob product is always (1/2)^4 = 1/16 no matter how many successes we're investigating. The one success in four tries combination is 4!/(1! x (4-1)1) = 4!/3! = 4. So for one rooster in four tries comes out as a 4 x 1/16 = 4/16 = 1/4 chance.
two roosters 4!/(2! x 2!) = 24/4 = 6 ways, so then 6/16 = 3/8 prob
three roosters 4!/(1! x 3!) = 4 ways, so then 4/16 = 1/4 prob (Notice the symmetry, since roosters and hens are assumed equally likely then three of one and one of the other doesn't care which is the one and which are the three.)
four roosters 4!/(0! x 4!) = 1 then 1/16
zero roosters is same as four hens by symmetry so zero roosters prob is also 1/16
So in terms of ordered pairs with (how many roosters, probability) we have
(0, 1/16), (1, 4/16), (2, 6/16), (3, 4/16), (4, 1/16)
If we sum over the probabilities we have 1/16 + 4/16 + 6/16 + 4/16 + 1/16 = 16/16 = 1, the whole probability enchilada.
Thanks, very succinct.
