A probability question for all you math lovers out there!?

Question

If a couple has the genetic make up making it a 4% chance that any child they have will inherit a particular disorder,
What are the odds of them having 4 children, two with the disorder.
How do you come to your figure?
My guess is .48%, or approx 5 out of a thousand.

How do you adjust your formula if a boy is 4 times more likely to inherit the condition than a girl 1% chance for a girl) & the only boy has the condition and one of the three girls has the condition.
My guess is on this is .18%, or approx 2 out of a thousand.

Answer 1

1. Using Binomial probability (0.96 + 0.04)^4

Probability (2 have and 2 have not) = 4C2x(0.96)²x(0.04)²
= 0.00885 (ie around 0.9% ... 9 in 1000)

P (boy has condition) = 0.04

Using Binomial probability (0.99 + 0.01)³
P (one girl has and 3 girls do not have condition) = 3C2x(0.99)²x(0.01)
=0.0294

So P (boy has condition AND one girl has and 3 girls do not have condition)
= P (boy has condition) * P(one girl has and 3 girls do not have condition)
= 0.04 x 0.0294
= 0.00118 (ie around 0.1% or 1 in 1000)

Answer 2

4C2 * 3C1 / 7C3 18/35 = .5143 you have a fifty one.40 3% danger of having an examination the place you recognize 2 out of the three questions you have a (4C0 * 3C3) / 7C3 = a million/35 = 2.86% danger of having an examination the place you recognize not one of the questions you have a 11.40 3% danger of having an examination the place you recognize all 3 questions you have a 34.29% danger of having an examination the place you recognize a million question