conditional probability of a family

Question

Family A has 6 members of which 4 are males and 2 are females & family B has 5 members of both genders.(totally 5 memebers) (Assume it is equi-probable for a member of unknown gender to be a male or female). 2 members are selected randomly either from family A or family B. If both members are female then probability that they belong to family A is?

I can understand that we have to use Bayes theorem here but here there are several cases for family B. How to solve in such cases.

Answer 1

if the probability to be M of F is equal $P(M)=P(F)=\frac{1}{2}$ any elements of the 30 events of the sample space are equiprobable, thus you have

$$P(F=k)=\frac{\binom{5}{k}}{\sum_{i=1}^{4}\binom{5}{i}=30}$$

$F=1,2,3,4$

Here is the sample space $\Omega$, for your help in understanding

Answer 2

Let $E_1$ and $E_2$ be the events that both members were picked from Family A and both members picked were female, respectively. We can use Bayes' Rule to compute $P(E_1 \mid E_2)$ as follows:

$$P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{P(E_1)P(E_2 \mid E_1)}{P(E_2)} .$$

To compute $P(E_1)$, the probability that both members were picked from Family A, we must note that each subset of members is equally likely. The desired probability is $P(E_1) = {6\choose 2}/{11\choose 2} = \frac{3}{11}$ since there are ${6\choose 2}$ possible subsets of two people from Family A out of ${11\choose 2}$ total possible subsets of two people.

Now we will compute $P(E_2 \mid E_1)$, the probability that both members picked were female given that both came from Family A. The required probability is $P(E_2 \mid E_1) = \frac{2}{6} \cdot \frac{1}{5} = \frac{1}{15}$ since it is just the probability of picking the two females without replacement.

Finally, we must compute $P(E_2)$, the probability of both members picked being female. This is a little bit trickier. Let's introduce a new random variable $X$ which denotes the number of females in Family B. Clearly, $X \sim \text{Bin}(5, 0.5)$. Now we can easily compute the result by conditioning on $X$. For a fixed value of $X = x$, the required answer is ${2 + x\choose 2}/{11\choose 2}$ since there are $2 + x$ total females (and we can choose any two of them) whereas there are $11$ total people (and we can choose any two of them when we don't have the constraint). Therefore,

$$P(E_2) = \sum_{x= 0}^{5} P(E_2 \mid X = x)P(X= x) = \frac{1}{55 \cdot 2^{5}}\left[{5\choose 0}{2\choose 2} + {5\choose 1}{3\choose 2} + {5\choose 2}{4\choose 2} + {5\choose 3}{5\choose 2} + {5\choose 4}{6\choose 2} + {5\choose 5}{7\choose 2}\right] = \frac{17}{110}.$$

Now we can just plug in the values to get the final answer:

$$P(E_1 \mid E_2) = \frac{\frac{3}{11} \cdot \frac{1}{15}}{\frac{17}{110}} = \boxed{\frac{2}{17}}$$

Answer 3

I have a different solution.

Assuming that the family B is composed by 5 members total (with both gender present and any guy with probability 0.5 to be male or female, independently), the distribution of # of females in B Family is

$$B_{Female} = \begin{cases} \frac{5}{30}, & \text{if $F=1$} \\ \frac{10}{30}, & \text{if $F=2$} \\ \frac{10}{30}, & \text{if $F=3$} \\ \frac{5}{30}, & \text{if $F=4$} \end{cases}$$

Thus the requested probabilty is

$$\mathbb{P}[A|FF]=\frac{\frac{1}{2}\cdot\frac{2}{6}\cdot\frac{1}{5}}{\frac{1}{2}\cdot\frac{2}{6}\cdot\frac{1}{5}+\frac{1}{2}[\frac{10}{30}\cdot\frac{2}{5}\cdot\frac{1}{4}+\frac{10}{30}\cdot\frac{3}{5}\cdot\frac{2}{4}+\frac{5}{30}\cdot\frac{4}{5}\cdot\frac{3}{4}]}=$$

...easy simplify $\frac{1}{60}$ in every addend...

$$=\frac{2}{2+[1+3+3]}=\frac{2}{9}$$