MHB Conditional probability: Eye color of sibling

AI Thread Summary
The discussion revolves around calculating the conditional probability of a second child being "bb" given that the first child is "bb." The initial calculations for the probability of the first child being "bb" are confirmed to be correct, leading to a probability of 1/4 for the second child also being "bb." However, confusion arises regarding the sample space and the branching of possibilities after knowing the first child's genotype. Ultimately, it is clarified that the correct probability for the second child remains 1/4, despite initial miscalculations. The conversation highlights the complexities of conditional probability in genetic scenarios.
carnivean1
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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 -.-
 
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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.
 
The course instructor corrected his answer, my solution is indeed correct...
 
carnivean said:
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!
 
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