98. Find the distribution series of the random variable $X$ - the number of times a six is rolled in three throws of a die. Calculate the mathematical expectation, variance, and standard deviation of this random variable.
S o l u t i o n. The experiment consists of one roll of a fair die. The total number of trials is $n=3$. The trials are independent, as the result of each previous trial does not affect the result of the subsequent one. In each trial, event $A$ - the appearance of a six - may or may not occur. The probability of event $A$ occurring in each trial is constant and equal to $p=1 / 6$, and the probability of event $A$ not occurring is $q=1-p=5 / 6$. The number of occurrences of event $A$ in $n$ trials is $m$; according to the condition, we have $m=0 ; 1 ; 2 ; 3$.

The random variable $X$ - the number of occurrences of event $A$ in three independent trials - can take one of the values $x_{i}=m_{i}(i=0 ; 1 ; 2 ; 3)$ as a result of the experiment. The probabilities of each of the possible values of $X$ will be found using the Bernoulli formula:

$$
\begin{aligned}
& p_{0}=P(X=0)=P_{3}(0)=C_{3}^{0} p^{0} q^{3}=1 \cdot\left(\frac{1}{6}\right)^{0} \cdot\left(\frac{5}{6}\right)^{3}=\frac{125}{216} \\
& p_{1}=P(X=1)=P_{3}(1)=C_{3}^{1} p^{1} q^{2}=3 \cdot\left(\frac{1}{6}\right)^{1} \cdot\left(\frac{5}{6}\right)^{2}=\frac{75}{216} \\
& p_{2}=P(X=2)=P_{3}(2)=C_{3}^{2} p^{2} q^{1}=3 \cdot\left(\frac{1}{6}\right)^{2} \cdot\left(\frac{5}{6}\right)^{1}=\frac{15}{216} \\
& p_{3}=P(X=3)=P_{3}(3)=C_{3}^{3} p^{3} q^{0}=1 \cdot\left(\frac{1}{6}\right)^{3} \cdot\left(\frac{5}{6}\right)^{0}=\frac{1}{216}
\end{aligned}
$$

The check, which involves finding the sum of the probabilities $p_{i}$, shows that this sum is equal to one: $125 / 216+75 / 216+15 / 216+1 / 216 \approx 1$.

The considered random variable has the following distribution series:

$$
X: \begin{array}{ccccc}
x_{i} & 0 & 1 & 2 & 3 \\
p_{i} & \frac{125}{216} & \frac{75}{216} & \frac{15}{216} & \frac{1}{216}
\end{array}
$$

The expected value, variance, and standard deviation of the variable $X$ will be calculated using formulas (44), (45), and (46):

$$
\begin{aligned}
& M(X)=n p=3 \cdot \frac{1}{6}=\frac{1}{2} ; D(X)=n p q=3 \cdot \frac{1}{6} \cdot \frac{5}{6}=\frac{5}{12} \\
& \sigma(X)=\sqrt{n p q}=\sqrt{\frac{5}{12}}=\frac{\sqrt{15}}{6}
\end{aligned}
$$