you got the first part correct..
for the second one:
At least 3 have blue eyes if and only if at least 2 of the other 4 have blue eyes. You can add the probabilities of these outcomes:
Hi all,
I'm looking for an answer to the following exercise. I've worked on some but I'm stuck. There is a family with 5 children, and the probability that any child will have blue eyes is 1/4. If you know that at least 1 has blue eyes, what is the probability that 3 or more have blue eyes? (I think I got this part, below.) The next part is: if you know that the youngest child in the family has blue eyes, what is the probability that 3 or more have blue eyes. Supposedly these answers are different.
I think for the first part you can say that the probability than n children have blue eyes is:
So the answer to the first part is, I think, a conditional probability:
The textbook has the answer for the second part as .25. But I don't see it. It seems to me you could just ignore the youngest child, and treat it now like a family of four and get the probability of 2, 3, or 4 children having blue eyes. But that's not .25, unless I'm calculating wrong (I get 67/256).
Thanks for any help, take care,
Rob