- #1

carnivean1

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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 -.-