215. Find the variance of a discrete random variable $X$ - the number of occurrences of event $A$ in two independent trials, if the probabilities of the event occurring in these trials are the same and it is known that $M(X)=1.2$.
Solution. First method. The possible values of the quantity $X$ are: $x_{1}=0$ (the event did not occur), $x_{2}=1$ (the event occurred once), and $x_{3}=2$ (the event occurred twice).

We will find the probabilities of the possible values using the Bernoulli formula: $P_{2}(0)=q^{2} ; \quad P_{2}(1)=C_{2}^{1} p q=2 p q ; \quad P_{2}(2)=p^{2}$.

We will write the distribution law of $X$:

$$
\begin{array}{llcc}
\text { possible values } & 0 & 1 & 2 \\
\text { probabilities } & q^{2} & 2 p q & p^{2}
\end{array}
$$

We will find $M(X):$

$$
M(X)=2 p q+2 p^{2}=2 p(q+p)=2 p
$$

By the condition $M(X)=1.2$, i.e., $2 p=1.2$. Hence, $p=0.6$ and, consequently, $q=1-0.6=0.4$.

The required variance

$$
D(X)=n p q=2 \cdot 0.6 \cdot 0.4=0.48
$$

Second method. We will use the formula $M(X)=n p$. By the condition, $M(X)=1.2 ; n=2$. Therefore, $1.2=2 p$. Hence, $p=0.6$ and, thus, $q=0.4$.

We will find the required variance:

$$
D(X)=n p q=2 \cdot 0.6 \cdot 0.4=0.48
$$

Of course, the second method leads to the goal more quickly.