# Thread:

1. ## Conditional probability - family with blue eyes

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:

$P_n = {5 \choose n}\frac14^n\frac34^{5-n}$

So the answer to the first part is, I think, a conditional probability:

${P_3 + P_4 + P_5 \over P_1+P_2+P_3 + P_4 + P_5} = \frac{106}{781}$

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

2. 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:

$6 \left(\frac{1}{4}\right)^{2} \left(\frac{3}{4}\right)^{2}+4\left(\frac{1}{4}\ri ght)^{3} \frac{3}{4}+\left(\frac{1}{4}\right)^4$

3. You're result is also 67/256. Perhaps this textbook (Degroot Probability and Statistics, 3rd ed) is wrong. The back of the book says .25.

4. As far as I know, it should be 0.26
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