Q.1) A family has $n$ children, $n\geq2$. We ask from the father, "Do you have at least one daughter named Lilia?" He replies, "Yes!". What is the probability that all of their children are girls?
In other words, we want to find the probability that all $n$ children are girls, given that the family has at least one daughter named Lilia.
Here we can assume that if a child is a girl, her name will be Lilia with probability $\alpha\ll1$ independently from other children's names. If the child is a boy, his name will not be Lilia.
Q.2) In a family of $n$ children. We pick one among them and found that she is a girl. What is the probability that all children are girls?
My solution to Q.1)
$$
\begin{equation}
\begin{split}
P(\text{all are girls | at-least one named Lila}) &= \frac{P(\text{at-least one name Lila | all are girls})\ \times\ P(\text{all are girls})}{P(\text{at-leat one named Lila})}\\
&= \frac{{n\choose1}\ \alpha\ (1-\alpha)^{n-1}\ \times\ \frac{1}{2^n}}{{n\choose1}\ \alpha \ \frac{1}{2^{n-1}}}
\end{split}
\end{equation}$$
My solution to Q.2)
$$\begin{equation}
\begin{split}
P(\text{all are girls | at-least one girl}) &= \frac{P(\text{at-least one girl | all are girls})\ \times\ P(\text{all are girls})}{P(\text{at-least one girl})}\\
&= \frac{1\ \times\ \frac{1}{2^n}}{{n\choose1}\ \frac{1}{2} \ \frac{1}{2^{n-1}}}
\end{split}
\end{equation}$$
Best Answer
Alright so the first question seems to be confusing some people. Let $B$ be the event of all boys. The complement of $B$ is $B^c$, which is the event of at least one girl. Here is where confusion then plays out: is $B^c$ the same as a girl named Lilia? Does this mean that if you have at least one girl, then her name will be Lilia? No because her name could have been any other name, the fact that the child is named Lilia simply indicates that in addition to having a child named Lilia, you now have at least one girl. Now let $L$ be the event that none of the $n$ children be named Lilia. There is some more hidden information. When the problem states that the probability of Lilia is $\alpha<<1$, the problem actually assumes that you are expected to see the names of other girls repeat, such as Mary, before the name Lilia is first encountered. That being said, let $L^c$ be the event of at least 1 Lilia. Let $G$ be the event that all children are girls, let $c_i$ denote the event of the $i_{th}$ child. Finally consider Baye's Rule: $$P(G|(B^c\cap{L^c})){\times}P(B^c\cap{L^c})= P((B^c\cap{L^c})|G){\times}P(G)$$ The problem specifically asks for $P(G|(B^c\cap{L^c}))$. So now it is time to find all the other pieces of equation. Assuming that child $c_i$ has an equal of probability of being girl or boy, then $P(G) = ({\frac{1}{2}})^n$. Moving on to the next term $P((B^c\cap{L^c})|G)$, which denotes the probability of at least one girl, and at least one Lilia, given that all the children are girls. If a child that is also a girl has a probability $\alpha$ of being named Lilia, then that child that is also a girl has a probability of $1-\alpha$ of not being named Lilia. $L$ is the event that none of the children are named Lilia, and in this particular case, there are n girls, and so now consider the probability that none of the n girls are named Lilia, which is $(1-\alpha)^n$. However, in reality, the event of at least one Lilia is sought, thus $P((B^c\cap{L^c})|G) = (1-(1-\alpha)^n)$. Here you can ignore the $B^c$, the event of at least one girl, because you are given n girls as per the conditional.
Finally consider the last portion of the problem, $P(B^c\cap L^c)$. This is the event of at least one girl, and at least one girl named Lilia. One basic rule in Set theory essentially allows this problem to be reformulated as $P((B \cup L)^c)$, which is the complement of the event of all boys or all children are not named Lilia. Then it follows that $P((B \cup L)^c) = 1 - P(B \cup L)$. To find how these quantities interact, consider the Addition Rule of non-disjoint sets such that $P(B \cup L) = P(B)+P(L)-P(B \cap L)$. The event of all boys is equal to $P(B) = (\frac{1}{2})^n$, and then $P(B \cap L) = P(L|B) \times P(B) = 1 \times (\frac{1}{2})^n$. This is true because if you have all boys, then probability of no Lilias is subsequently 1. This means that $P(B \cup L) = P(L)$. So what is the probability that no child will be named Lilia. Is important to consider that this includes both boys and girls, because boys will certainly not be named Lilia, and only a select number of girls will be named Lilia. Let $L_i$ be the event where the $i_{th}$ child is not named Lilia. Then $$P(L_i) = P(L_i \cap c_i=g)+P(L_i \cap c_i=b) = P(L_i|c_i=g) \times P(c_i=g) + P(L_i|c_i=b) \times P(c_i=b)$$ This yields, $P(L_i) = (1-\alpha) \times (\frac{1}{2})+(1) \times (\frac{1}{2}) = (\frac{1}{2})(2-\alpha)$. Then the probability that all n children are not named Lilia is $P(L) = ((\frac{1}{2})(2-\alpha))^n = \frac{(2-\alpha)^n}{2^n}$. Going a little far back, $P((B \cup L)^c) = 1-P(L) = 1-\frac{(2-\alpha)^n}{2^n}= \frac{2^n-(2-\alpha)^n}{2^n} $
Finally, $$P(G|(B^c\cap{L^c})) = \frac{P((B^c\cap{L^c})|G){\times}P(G)}{P(B^c\cap{L^c})}$$
$$= ((1-(1-\alpha)^n) \times \frac{1}{2^n}) \div \frac{2^n-(2-\alpha)^n}{2^n}$$
$$= (\frac{((1-(1-\alpha)^n)}{2^n}) \times (\frac{2^n}{2^n-(2-\alpha)^n}$$
Obviously the answer is completely dependent on being given a girl named Lilia. It might not seem intuitive at first, but consider it this way, instead of boys and girls, use blue and red colored shapes. Blue shapes have an odd number, and red shapes have an even number. Also, imagine having way more lower numbers than larger numbers. So if I tell you that there is at least one red colored shape with a really large even number, then the predicted number of reds should go up from your initial estimate.
Problem 2 Okay and for problem 2, I am assuming that a girl is picked at random. If that's the case, then the problem is quickly reduced. First let, $E$ be the event of random picking a girl. However this isn't all that is actually said. In reality, you picked one of the children, and then it turns out that the child may be either boy or girl. Let $p_i$ be the event of selecting $i_th$ child. This is all equivalent to $E = (C_1 \cap p_1) \cup (C_2 \cap p_2) \cup ... \cup (C_n \cap p_n)$. Note that each of these events are actually disjoint, that is picking the 1st child is separate from picking the 2nd child and so on. Thus $$P(E) = P(C_1 \cap p_1) +P(C_2 \cap p_2) + ... +P(C_n \cap p_n) = P(C_1 | p_1) \times P(p_1) + P(C_2 | p_2) \times P(p_2) + ... + P(C_n | p_n) \times P(p_n) $$ Thus, $$P(E) = (\frac{1}{2} \times \frac{1}{n}) + (\frac{1}{2} \times \frac{1}{n}) + ... + (\frac{1}{2} \times \frac{1}{n}) = (\frac{1}{2n}) \times (1+1...+1) = \frac{n}{2n} = \frac{1}{2}$$
Now use Bayes Rule to solve the remainder of the problem. Let $G$ be the event that all children are girls. So $$P(G | (B^c \cap E)) \times P(B^c \cap E) = P((B^c \cap E) | G) \times P(G) = P(G | (B^c \cap E)) \times P(B^c | E) \times P(E) = 1 \times \frac{1}{2^n}$$ So if you are given all girls, the probability of at least 1 girl AND picking a girl at random is now 1. Hopefully the probability of G is obvious. It should also be obvious that if you pick any girl at random, then you have at least 1 girl so that is simply 1. Then, $$P(G | (B^c \cap E)) = (\frac{1}{2^n}) \div (1 \times \frac{1}{2}) = (\frac{1}{2^n}) \times 2 = \frac{1}{2^{n-1}}$$ A similar but more complicated approach can be reached using the binomial expansion although it can be tedious to follow along.