In the paternity suit, the mother's blood type is A, the man's blood type is B, and the child's blood type is AB. According to various circumstances, the possibility that the man identified as a real father is 50:50. Based on the given blood type data, find the probability that he is a real father, if the father's blood type is A or O (86% of the total population), the child cannot be AB, and if the father's blood type is B or AB (14% of the total population), the probability that the child is AB is 1/4. If the man is not the father, the father is considered the representative of the general population.
I'm trying to solve it with Bayes Theorem. Below are the events I defined.
P(F): Event that he is a real father. We know that the prior is 1/2 by the second paragraph.
P(F$^c$) : Complement of P(F) which is 1/2 as well
P(AB|F) : Given that he is a real father, probability of the child's blood type is AB
P(AB|F$^c$) : Given that he is not a real father, probability of the child's blood type is AB
P(F|AB) : Posterior that I would like to know.
P(F|AB) = $\frac{P(AB|F)P(F)}{P(AB|F)P(F) + P(AB|F^c)P(F^c)}$
Here's the thing, I'm not sure about P(AB|F$^c$). Did I define the events wrong or Did I miss something...(I don't think I understand the meaning of the last paragraph => "If the man is not the father, the father is considered the representative of the general population")
Can anyone help me with this?
Thank you.
In a (convoluted) Bayes Theorem problem like this, I try to attack the problem in pieces, rather than set up one complicated fraction.
Let $E_1$ denote the event that the man is the father.
Let $E_2$ denote the event that the child's blood type is AB.
You want
$$p(E_1 | E_2) = \frac{p(E_1,E_2)}{p(E_2)}. \tag1 $$
So the problem has been reduced to computing the probability of two separate situations:
$\underline{\text{Computing the probability of} ~(E_1,E_2)}$
Absent any other information, you know that $p(E_1) = (1/2)$.
Further, you know that $p(E_2|E_1)$, which represents the event that the child is AB, given that the father is type B, is $(1/4)$.
Therefore,
$$p(E_1,E_2) = p(E_1) \times p(E_2|E_1) = (1/2) \times (1/4) = (1/8).$$
$\underline{\text{Computing the probability of} ~E_2}$
In order to compute $p(E_2)$, you have to consider two mutually exclusive situations, and add up the probabilities of each situation:
Either the man is the father : $~p(E_2,E_1)$
Or the man is not the father : $~p(E_2,\neg E_1).$
From the previous section, you have that $p(E_2,E_1) = (1/8).$
To compute $p(E_2,\neg E_1),$ consider that this requires three things to happen:
The man is not the father : probability is $(1/2)$
The father, being from the general population, is B or AB : probability is $\frac{14}{100}$.
The child then being AB : probability is $(1/4)$.
$\displaystyle \frac{1}{2} \times \frac{14}{100} \times \frac{1}{4} = \frac{7}{400}.$
Therefore,
$$p(E_2) = \frac{1}{8} + \frac{7}{400} = \frac{57}{400}.$$
$\underline{\text{Final computation}}$
$$p(E_1|E_2) = \frac{p(E_1,E_2)}{p(E_2)}$$
$$= \frac{\frac{1}{8}}{\frac{57}{400}} = \frac{400}{8 \times 57} = \frac{50}{57}.$$
Addendum
There is a sophisticated shortcut, that I should mention, even though I don't recommend it.
Absent any other information, the probability that the man is the father is $(1/2)$. So, when comparing the relative probabilities of the man being or not being the father, absent any other information, the factor of $(1/2)$ is neutral.
If the father (is the man, or someone else with blood type B or AB), the probability of the son having blood type AB is $(1/4)$. This factor also applies to the two situations of the father being the man, or someone else.
So, the only pertinent information, when deciding the relative probabilities of the father being the man or someone else is that the man is guaranteed to have a satisfactory blood type (AB or B), while someone else will only have a satisfactory blood type with probability $\dfrac{14}{100}.$
Therefore, you can reason that the odds that the man is the father must be
$$\frac{100}{100} ~~\text{to}~~ \frac{14}{100}.$$
This translates to the probability of the man being the father equaling
$$\frac{100}{100 + 14}.$$