Binomial Question

In the chapter we are given the formula P(X=r) = nCr * q^n-r * p^r

1) A recovery ward is a maternity hospital has 6 beds. What is the probability that the mothers there have between them 4 girls and 2 boys?
(Assuming there are no twins and a baby is equally likely to be a girl or boy)

I did n = 6 , r = 2 boys
p = probability of girl = 4/6 = 2/3
q= probability of boy = 2/6 = 1/3

so P(X=2) = 6C2 x (1/3)^6-2 x (2/3)^2
= 6C2 x (1/3)^4 x (2/3)^2
=0.08 or 20/243
however the answer says 15/64 so I don't know how to do it, please help!

It is not dependent on the numbers 4,2. They are represented by the exponents in the binomial distribution.

ohh thanks, so p and q are 0.5? how do I know what r is?

You could use r=2 or 4 as p=q=0.5.

ohh okay thank you

if boy / girl are equally likely, what is p?
It is not dependent on the numbers 4,2. They are represented by the exponents in the binomial distribution.