15. Random Vectors. There are $n$ random vectors of the form $\left(y_{1}, y_{2}, y_{3}\right)$, where exactly one random coordinate is equal to 1, and the others are 0. They are added together. The result is a random vector $\vec{a}$ with coordinates $\left(Y_{1}, Y_{2}, Y_{3}\right)$.

a) (from 9th grade. 2 points). Find the expected value of the random variable $\vec{a}^{2}$.

b) (from 9th grade. 2 points). Prove that $|\vec{a}| \geq \frac{n}{\sqrt{3}}$.
Solution. a) Let's find the expected value $\mathrm{E} Y_{j}^{2}$. The quantity $Y_{j}$ is the number of ones among the numbers $y_{j}$, that is, this quantity is distributed according to the binomial law with probability $p=1 / 3$ and the number of trials $n$. Therefore,

$$
\mathrm{E} Y_{j}=n \cdot \frac{1}{3}=\frac{n}{3} \text { and } \mathrm{D} Y_{j}=n \cdot \frac{1}{3} \cdot \frac{2}{3}=\frac{2 n}{9}
$$

Then

$$
\mathrm{E}\left(\vec{a}^{2}\right)=\mathrm{E}\left(Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}\right)=3 \mathrm{E} Y_{1}^{2}=3\left(\mathrm{D} Y_{1}+\mathrm{E}^{2} Y_{1}\right)=3 \cdot \frac{2 n}{9}+3 \cdot \frac{n^{2}}{9}=\frac{2 n+n^{2}}{3}
$$

b) We will use a known fact: the root mean square of non-negative numbers is not less than their arithmetic mean.

$$
\sqrt{Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}}=\sqrt{\frac{Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}}{3}} \cdot \sqrt{3} \geq \frac{Y_{1}+Y_{2}+Y_{3}}{3} \cdot \sqrt{3}=\frac{n}{\sqrt{3}}
$$

Answer: a) $\frac{2 n+n^{2}}{3}$.