Task 1. The random variable $X$ is given by the probability density function $f(x)=a x^{2}$ for $x \in(0 ; 1)$ and $f(x)=0$ for $x \notin(0 ; 1)$. Determine the parameter $a$ and the distribution function $F(x)$. Also, find the probability that in four trials, $X$ will exactly three times take values in the interval $(0,25 ; 0,75)$.
Solution. The value of the parameter $a$ is determined from the condition $\int_{-\infty}^{+\infty} f(x) d x=1$. We have $\int_{0}^{1} a x^{2} d x=\left.\frac{a x^{3}}{3}\right|_{0} ^{1}=\frac{a}{3}=1$. We considered that $f(x)=0$ for $x \notin(0 ; 1)$. Hence, $a=3$. Thus, we have $f(x)=3 x^{2}, x \in(0 ; 1)$ and $f(x)=0, x \notin(0 ; 1)$ (Fig. 7.10, a). Next, $F(x)=\int_{-\infty}^{x} f(t) d t$. If $x \leqslant 0$, then $F(x)=0$, since $f(x) \equiv 0$; if $x \in(0 ; 1)$, then $F(x)=\int_{0}^{x} 3 t^{2} d t=\left.t^{3}\right|_{0} ^{x}=x^{3}$, and if $x>1$, then $F(x)=\int_{0}^{1} 3 x^{2} d x=\left.x^{3}\right|_{0} ^{1}=1$ (Fig. 7.10, b). Thus,

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-157.jpg?height=288&width=874&top_left_y=176&top_left_x=1482)

Fig 710

$$
f(x)=\left[\begin{array}{ll}
0, & x \leqslant 0, x \geqslant 1 ; \\
3 x^{2}, & 0<x<1 ;
\end{array} \quad F(x)=\left[\begin{array}{ll}
0, & x \leqslant 0 \\
x^{3}, & 0<x<1 \\
1, & x \geqslant 1
\end{array}\right.\right.
$$

Now, let's determine $p(0.25 < x \leqslant 0.75) = F(0.75) - F(0.25) = 0.421875 - 0.015625 \approx 0.4$.

Finally, using the Bernoulli formula (see Chapter VI, §7), we compute

$$
p_{4}(3)=C_{4}^{3}(0.4)^{3} 0.6=4 \cdot 0.064 \cdot 0.6=0.1536
$$