15. Twins. The probability of twins being born in Shvambria is $p$, and triplets are not born in Shvambria.

a) Estimate the probability that a Shvambrian met on the street is one of a pair of twins?

b) In a certain Shvambrian family, there are three children. What is the probability that among them there is a pair of twins?

c) In Shvambrian schools, twins are always enrolled in the same class. There are a total of $N$ first-graders in Shvambria. What is the expected number of pairs of twins among them?
Solution. a) Let $M$ pairs of twins and $N$ "singletons" walk the streets of the Shvambrian towns and villages. Then $\frac{M}{M+N} \approx p$, from which $\frac{M}{N} \approx \frac{p}{1-p}$. If we meet a random Shvambrian, the probability that they are one of the twins is

$$
\frac{2 M}{2 M+N}=\frac{2 \frac{M}{N}}{2 \frac{M}{N}+1} \approx \frac{2 \frac{p}{1-p}}{2 \frac{p}{1-p}+1}=\frac{2 p}{p+1}
$$

b) The desired probability is conditional. To understand this, let's draw a graph. Red edges indicate the birth of twins. Blue edges indicate singletons. The graph can be extended indefinitely downward, but we will only draw the fragment of the graph that accounts for all possible ways of having three children. We will label each edge with its probability. We will represent the event $A=\{$ Three children $\}$ with a curved area. This event includes three vertices, i.e., three elementary events with a total probability of $2 p q+q^{3}$.

Of these, the event $B=\{$ There are twins $\}$ is favorable in two elementary events (chains containing red edges) with a total probability of $2 p q$. Then the conditional probability is

$$
\mathrm{P}(B \mid A)=\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)}=\frac{2 p q}{2 p q+q^{3}}=\frac{2 p}{2 p+q^{2}}
$$

c) As we know, the probability that a randomly chosen first-grader is one of a pair of twins is $\frac{2 p}{p+1}$. Let's introduce a random variable $I_{k}$ - the indicator of a twin. $I_{k}=1$ if the $k$-th first-grader is one of a pair of twins and $I_{k}=0$ if the $k$-th first-grader is a singleton.

$$
\mathrm{E} I_{k}=1 \cdot \frac{2 p}{p+1}+0 \cdot \frac{1-p}{p+1}=\frac{2 p}{p+1}
$$

Then the total number of pairs of first-graders who are twins is

$$
S=\frac{1}{2}\left(I_{1}+I_{2}+\ldots+I_{N}\right)
$$

Transition to expectations:

$$
\mathrm{E} S=\frac{1}{2}\left(\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{N}\right)=\frac{1}{2} \cdot N \cdot \frac{2 p}{p+1}=\frac{N p}{p+1}
$$

Answer: a) $\left.\frac{2 p}{1+p} ; \boldsymbol{\sigma}\right) \frac{2 p}{2 p+q^{2}} ; \mathbf{c )} \frac{N p}{p+1}$