Conditional probability: Eye color of sibling

[carnivean1]

Ok I am really stumped and have no clue what I am missing...Here is the scenario. We have a mother and father 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 chosen.
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 $bb$ 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 -.-

 

[Jameson]

Hi carnivean,

Welcome to MHB! :)

These questions are really confusing and often involve counter intuitive solutions. If you can check your answer then I might try this (although I'm not 100% sure it's correct).

There are 16 combinations (or $2^4$) we can make with $B,B,b,b$. We only care about the ones that start with $bb$ though, since the first child is known to be $bb$. So if we start with four branches for the first child and then branch off from that, how many possibilities do we have? We have two levels basically... 4 branches for the first child and 4 more off each of those. I think the answer should be 1/4 again because only one branch off our original $bb$ branch satisfies the condition.

Sometimes this will be worded like "at least one child is $bb$", in which case we have a different sample space. I would try 1/4 though if you can check your answer. I'm curious if this is correct.

 

[carnivean1]

The course instructor corrected his answer, my solution is indeed correct...

 

[Jameson]

The course instructor corrected his answer, my solution is indeed correct...

Ok, good to know. I didn't see anything wrong with your logic but was trying to think of it differently. I believe my method doesn't work because it's a second generation problem - we have the parents genes that are unknown who pass them along to kids. I was referring to questions like you see here.

Thank you for the followup!