Problem, calculating the conditional Probability I am preparing for my exam and have come across this task that causes me problems.
A disease is genetic, and occurs when both genes have a particular mutation. If the disease occurs, the person dies before it can reproduce, that is, as a child. We use the following model. A child receives a gene from father and mother independently of each other. Parents are equally likely to give each of their two genes on. In the adult population, regardless of gender, the percentage of people who do not carry a mutated gene is $\alpha$. Pair formation takes place
regardless of whether the persons carry a mutated gene or not.
a) Calculate the probability that a child is ill and the probability,
that a child doesn't have a mutated gene.
b) Calculate the proportion of the next generation adult population,
that doesn't carry a mutated gene.
c) We assume that the disease has existed for a very long time, and that
so that the portion $\alpha \in$ [0; 1] is stable, i. e. it does not change in the next Generation changes. Calculate $ \alpha $ .
d) Interpret the result from c).
I've done this assignment and I'm gonna get this: 
a): Define the sets: 
$$M_f = \textrm{Father has mutated gen}\\
\overline{M_f} = \textrm{Father has no mutated gen}\\ 
M_m = \textrm{Mother has mutated gen}\\\
\overline{M_m} =  \textrm{Mother has no mutated gen}\\ M_c= \textrm{Children has mutated gen}\\
\overline{M_c} = \textrm{children has no mutated gen}\\ \textrm{Then:}\\ P[M_c] = \frac{1}{2} P[M_m] \times \frac{1}{2} P[M_f] = \frac{1}{4} (1 - \alpha)^2 \\ \textrm{and} \\ P[\overline{M_c}] = \frac{1}{2} P[\overline{M_m}] \times \frac{1}{2} P[\overline{M_f}] = \frac{1}{4} \alpha^2 $$
The first solution is right, but the second is wrong. The solutions say that the result for the second one must be $ \frac{1}{4} (1+\alpha)^2$. That is why i stopped working on this exercise. Can anyone tell me how they got the result and why my result is wrong? Could someone also tell me how to go an on b)?
thank you in advance! 
I try to use the formula down in the comment:
$$ P[\overline{M_c}] = P[\overline{M_c} \vert M_v M_f] P[M_v M_f] + P[\overline{M_c} \vert \overline{M_v} M_f] P[ \overline{M_v }M_f] + P[\overline{M_c} \vert \overline{M_f} M_v] P[ \overline{M_f }M_v] +P[\overline{M_c} \vert \overline{M_v} M_f] P[ \overline{M_v } \overline {M_f}] $$
i would say that $$P[M_v M_f] = \frac{1}{2} \alpha * \frac{1}{2} \alpha = \frac{1}{4} \alpha^2 \\
P[ \overline{M_v }M_f] = \frac{1}{2} \alpha * \frac{1}{2} (1 - \alpha) = \frac{1}{4} \alpha *(1- \alpha) = P[ \overline{M_f }M_v] \\
 P[ \overline{M_v } \overline {M_f}] = \frac{1}{4} (1-\alpha)^2$$
Can someone tell me, how to calculate the probability of $$ P[\overline{M_c} \vert M_v M_f] \\
P[\overline{M_c} \vert \overline{M_v} M_f]\\
... $$
 A: There are 3 ways 
1)both parents do not have mutated genes:probability=(alpha)^2
Probability that no mutated gene transported to child and both parents are normal is (alpha)^2(both genes are normal any case is favourable)
2)both patents have mutated genes :probability (1-alpha)^2
Probability that no mutated gene transported to child and both parents are carriers =1/4(1-alpha)^2(its equally probable from both parents to give normal gene so a factor 1/4)
3)one parent is normal and one parent is carrier:probability = 2*(1-alpha)(alpha)
(Now why the extra factor 2, this can be easily viewed though either mother is carrier or father is carrier. Take a example of deck of 52 cards , now tell me which is more probable , getting 4 aces or 1ace, 1 queen, 1king  and 1 jack .if you have any doubts simply do it by: 1-(alpha)^2 -(1-alpha)^2)
Now probability of no mutated gene transported and one parent carrier and one parent normal=1/2*2*(1-alpha)(alpha).  (An extra factor 1/2 appeared because one parent who is normal can give any of the genes but the other parent have to give one of 2 possibe gene)
Therefore total probability(addition of above three cases)
=(Alpha)^2 + 1/4*(1-alpha)^2 + (alpha)*(1-alpha)
=1/4*(1 + (alpha)^2 + 2*(alpha) ) = 1/4* (1+ alpha)^2
I think u can figure out your mistake
