I'm going through Wasserman's All of Statistics. One of the questions in the first chapter is…
The probability that a child has blue eyes is 1/4. Assume independence
between children. Consider a family with 3 children.
(a) If it is known that at least one child has blue eyes, what is the
probability that at least two children have blue eyes?
The classic way of doing this question is simply to calculate:
$P(AtLeast1blue) = 1 – 27/64$
$P(AtLeast2Blue) = 9/64 + 1/64$
Then you just divide the two above to get
$\frac{10}{37}$
Here's what I don't quite understand. Given that we know at least 1 child has blue eyes, why can't we just calculate the probability that at least one of the two remaining children have blue eyes and have that be our answer?
So why is this incorrect: $\frac{1}{4}$ * $\frac{3}{4}$* $2$ + $\frac{1}{4}$*$\frac{1}{4}$ = $\frac{7}{16}$
The logic again, is just that we know one child has blue eyes so we take that as 100%. Then we just figure out the probability at least 1 of the remaining two children have blue eyes and mulitliy all these things together.
There's obviously some insight into the nature of probability and conditional probabilities that I'm missing, but I can't quite get it. It's probably something obvious, but I can't quite figure out why the above is wrong.
Best Answer
How are you making the choice of "one from the at least one" and which are "the remaining" children? Moreover, when you select these "remaining children", why would they each have the unbiased population probability (1/4) for having blue eyes? You are selecting from a biased population and that bias is the very thing you need to determine.