Ok I am really stumped and have no clue what I am missing…
Here is the scenario. We have a mother and farther who can be either BB, Bb or bb.
The probability for both is:
$$P(BB) = P(bb) = \frac{1}{4}$$
$$P(Bb) = \frac{1}{2}$$
For the child one gene of the mother and one from the father is randomly choosen.
We want to compute the chance for a child to be $bb$.
Let us denote the parents with:
$$P(mother, father)$$
So the sample space is (- i only posted the part of the sample space that we need not the whole):
$$P(bb,bb) = (\frac{1}{4})^2=\frac{1}{16}$$
$$P(Bb,bb) = P(bb,Bb) =\frac{1}{4} \frac{1}{2}=\frac{1}{8}$$
$$P(Bb,Bb) = (\frac{1}{2})^2=\frac{1}{4}$$
So $P(child = bb) $ is as follows:
$$ P(bb,bb) * 1 + P(Bb,bb) * 0.5 + P(bb,Bb) * 0.5 + P(Bb,Bb) * 0.25$$
$$ = 4 * \frac{1}{16} = \frac {1}{4}$$
This part is correct.
In the next part we know, that the first child of a family has $bb$ and we have to calculate the chance that the 2nd child has blue eyes as well.
My reasoning:
Each path gives us a chance of $\frac{1}{16}$ for the first child being $bb$, so if we know that, we have the following sample space for the parents:
$$(bb,bb), (Bb,bb), (bb,Bb), (Bb, Bb)$$ each is equally likely to be the parent combination.
$$P(bb,bb) = P(Bb,bb) = P(bb,Bb) = P(Bb, Bb) = \frac{1}{4}$$
The chance of the 2nd child being also $bb$ is:
$$ P(bb,bb) * 1 + P(Bb,bb) * 0.5 + P(bb,Bb) * 0.5 + P(Bb,Bb) * 0.25$$
$$ = \frac{1}{4} + \frac{1}{4}*0.5+ \frac{1}{4}*0.5 + \frac{1}{4}*0.25 = \frac {9}{16}$$
But this answer is wrong and I don't know why, could someone please help, I even simulated it and got the exact same result -.-
Best Answer
Maybe if we back up a step and think about the second question as we get the first. So what are the odds each set of parents will have two blue-eyed kids in a row?
P(bb,bb) * 1 * 1 + P(Bb,bb) * 0.5 * 0.5 + P(bb,Bb) * 0.5 *0.5 + P(Bb,Bb) ∗ 0.25 *0.25 = 9/64
What's the correct answer? Perhaps we can work backwards.