Bloody Problem

it is easier to work with ratio of alleles (gene). let the ratio of alleles $A,B,O$ be $\alpha,\beta,\omega$ . every new born receive 2 random alleles from this gene pool. so the probability of newborn of each genotype is $\begin{aligned} P(OO) &= \omega^2 \\ P(AO) &= 2 \alpha \omega \\ P(BO) &= 2 \beta \omega \\ P(AA) &= \alpha^2 \\ P(BB) &= \beta^2 \\ P(AB) &= 2 \alpha \beta \\ \text{total} &= (\omega + \alpha + \beta)^2 = 1 \end{aligned}$

since the ratio is "stable", the probabilities of newborn with blood type $A, AB$ are $0.32, 0.04$ respectively. $\begin{aligned} P(A) = P(AA) + P(AO) = \alpha^2 + 2 \alpha \omega &= 0.32 --(1) \\ P(AB) = \quad \quad \ 2 \alpha \beta &= 0.04 --(2) \\ \alpha + \beta + \omega &= 1.00 --(3) \\ \\ 2 \alpha \times (3) - (2) - (1) , \quad \alpha^2 &= 2 \alpha - 0.36 \\ \alpha &= 0.2 \text{ or } 1.8 \text{ (rejected) } \\ \beta &= 0.1 \\ \omega &= 0.7 \\ P(OO) &= 0.49 \\ \therefore 49 \text{\%} \end{aligned}$

We've been given the following information:

$\begin{aligned} &P(AA)+P(AO) = 0.32 = \frac{8}{25} \\ &P(AB) = 0.04 = \frac{1}{25}\end{aligned}$

Assuming that each person carries two alleles (allele is a variant of of a certain gene) and that the ratio of genotypes is stable , the probability that person has genotype $XY$ is equal to the product of the number of ways to arrange (XY) and probabilities of picking out X and Y, respectively. Assuming the stable ratio, probability of blood type $AA$ is:

$P(AA) = \left(P(AA) + \frac{P(AO)}{2} + \frac{P(AB)}{2}\right)^{2}$

By substituting $P(AO)$ for $\frac{8}{25} - P(AA)$ , we get quadratic equation of $P(AA)$ with solutions $\frac{1}{25}$ and $\frac{81}{25}$ , but we dismiss the latter because it's greater than 1. So, $P(AA) = \frac{1}{25}.$

Next, we write out the equation for $P(AB)$ :

$\begin{aligned} P(AB) &= 2\cdot \left(P(AA) + \frac{P(AO)}{2} + \frac{P(AB)}{2}\right)\cdot \left(P(BB) + \frac{P(BO)}{2} + \frac{P(AB)}{2}\right) \\ \frac{4}{50} &= P(BB) + \frac{P(BO)}{2}. \end{aligned}$

Then, we write out the equation for $P(BB)$ :

$\begin{aligned} P(BB) &= \left(P(BB) + \frac{P(BO)}{2} + \frac{P(AB)}{2}\right)^{2} \\ &= \left(\frac{4}{50} + \frac{1}{50}\right)^{2} = \frac{1}{100}\end{aligned}$

It follows that $BO = \dfrac{7}{50}$ , and finally $P(OO) = 1 - \left(\dfrac{8}{25} + \dfrac{1}{25} + \dfrac{7}{50} + \dfrac{1}{100}\right) = \dfrac{49}{100} = \boxed{49\%}.$

Note : One can show that this method is equivalent to @Albert Lau 's except it is carried out using ratios of blood types instead of alleles. For example, one can show that $P(OO) = \omega^{2}$ , where $\omega$ is the frequency of allele $O$ . By definition:

$\omega = \dfrac{\text{\# of O alleles}}{\text{total \# of alleles}} = \dfrac{2P(OO)\cdot N + P(AO)\cdot N + P(BO)\cdot N}{2N} = P(OO) + \dfrac{P(AO)}{2} + \dfrac{P(BO)}{2}.$

Hence, $P(OO) = \left(P(OO) + \frac{P(AO)}{2} + \frac{P(BO)}{2}\right)^{2} = \omega^{2}$