Question

a disease is inherited by a child wit probability 1/4. in a family of 2 children the prob that exactly one sibling is affected by the disease is?

Answer 1

The probabilities are not related - that is, whether or not the first child has the disease has absolutely no effect on whether the second child will have the disease as well.

So the probability that the first child has the disease is (1/4). The probability that the second child had the disease is also (1/4).

To get the probability that at least one of the kids will have the disease, you add the probabilities - so 1/4 + 1/4 = 1/2.

Answer 2

Oh, I misread the question. I read it as "at least" one - whereas it actually says "exactly one." I'm sorry - but it would have been nicer if you'd made the options available to the tutors. :)

Answer 3

Basically it changes the problem in that you still have two independent cases (detailed below) - but within each of those single cases, the two events (that child 1 and child 2 have the disease) are no longer independent.

So case 1:

The first child has the disease with probability (1/4) and the second child does not have it with probability (3/4). So the probability that both these events happen is the product of the probabilities (1/4)*(3/4) = 3/16

Case 2:

The first child does not have the disease with probability (3/4) and the second child does have with probability (1/4). So the probability of these two conditional events is (1/4)(3/4) = 3/16.

And to put the two cases together, the probability that one of them comes true is

3/16 + 3/16 = 6/16.

Which if you simplify is 3/8.