Explain the calculation for new this simple probability problem (ANSWER GIVEN)?

Question

Two heterozygous parents for albinism. If they have four children, what is the probability of them having 3 normal children and one albino?

For people without biological knowledge, the probability of having a normal child will be 3/4, the probability of having an albino child is 1/4. The answer to this question is 27/64, which happens to be (3/4)^3, or also (3/4)^3(1/4) x 4C3, but honestly i have no idea what is the correct way to think about or get to the answer, i just had the answer and noticed those two ways to get to it. please explain work, thank you

Answer 1

For each child, the probability of being normal is 3/4 and the probability of being albino is 1/4. So we are doing four independent trials of this experiment (four children) with identical probability of albinism for each: it's a binomial probability situation.

The binomial probability formula says that if the probability of success in each trial is p, the probability of exactly k successes in n trials is
(nCk) p^k (1-p)^(n-k)
Here we define "success" as albinism, so p = 1/4 and k = 1 and we get
(4C1) (1/4)^1 (3/4)^3 = 27/64.

Note that if we define "success" as normality, then p = 3/4 and we want k = 3, so we get
(4C3) (3/4)^3 (1/4)^1 = 27/64.
It's straightforward to show that the probability is always the same if you swap the definitions of success and failure and update the other parameters accordingly - as we should expect.

Answer 2

You have four children, so you calculate the probability.
The first is normal, so that's 3/4
Same with the second and third.
The fourth child is albino, with a 1/4 chance.
You multiply these because within the 3/4 chance of the first child, there's a 3/4 chance of another normal child and so on.
It comes out to 27/64