Distribution Question Stat In a particular town 10% of the families have no children, 20% have one child, 40% have two children, 20% have three children, and 10% have four. Let T represent the total number of children, and G the number of girls, in a family chosen at random from this town. Assuming that children are equally likely to be boys or girls, find the distribution of G. Please Display your answer in a table.
I tried, this but my set up didn't help me: P(G = n| G = B = 1/2) where n is the number of girls, and probability of G(Girl) = B(boy) = 1/2. I used conditional probabilities, but didn't help me, setup? Also, I can say that the distribution when G = 0 is 0. How to incorporate the number of childrens total? I only used the fact that we are interested in the number of girls, and boy and girl is equally likely
 A: I will begin the calculation, and you can complete it.  The random variable $G$ (the number of girls) can take on no values other than $0$, $1$, $2$, $3$, and $4$.  The easiest thing to calculate is $P(G=4)$.
How can there be $4$ girls? We must be dealing with a family of $4$ children (probability $0.1$) and that family must have $4$ girls.  The probability of $4$ girls given that there are $4$ children is $1/2^4$.  So the probability that a family has $4$ children and they are all girls is $(0.1)(1/16)$. To prepare for later, let's do this again with symbols.
$$P((T=4)\cap (G=4))=P((G=4)|(T=4))P(T=4)=(1/2^4)(0.1)$$
Now let us calculate the probability that $G=3$. This event can happen in two ways: (a) The family has $3$ children and they are all girls OR (b) the family has $4$ children and exactly $3$ of them are girls. 
First tackle (a). Let us find the probability of $3$ children, all girls. The probability of $3$ children is $0.2$. Given that a family has $3$ children, the probability they are all girls is $1/8$.  So the probability of $3$ children and all girls is $(0.2)(1/8)$. Now tackle (b). The probability of $4$ children is $(0.1)$. Given that there are $4$ children, the probability there are exactly $3$ girls is $4/16$ (I hope this part is known to you).  So the probability of $4$ children and exactly $3$ girls is $(0.1)(4/16)$. Add up the two probabilities we obtained. The probability of $3$ girls is $(0.2)(1/8)+(0.1)(4/16)$. In symbols,
$$P(G=3)= P((G=3)|(T=3))P(T=3)+P((G=3)|(T=4))P(T=4)$$
Now you can proceed to find the probability of $2$ girls ($3$ cases) also $1$ girl, also $0$ girls.  If you have calculated $4$ probabilities, the fifth can be found by subtracting the sum of the others from $1$.  But you might as well calculate all $5$ probabilities, and use the fact they must add up to $1$ as a check of your calculations.
