4. A random variable $\xi$ is uniformly distributed on the interval $[0 ; 6]$. Find the probability that the inequality $x^{2}+(2 \xi+1) x+3-\xi \leq 0$ holds for all $x \in[-2 ;-1]$.
Solution. The inequality $f(x)=x^{2}+(2 \xi+1) x+3-\xi \leq 0$ holds on the interval $[-2 ;-1]$, if

$$
\left\{\begin{array} { l } 
{ f ( - 1 ) \leq 0 } \\
{ f ( - 2 ) \leq 0 }
\end{array} \rightarrow \left\{\begin{array} { c } 
{ 1 - ( 2 \xi + 1 ) + 3 - \xi \leq 0 } \\
{ 1 - 2 ( 2 \xi + 1 ) + 3 - \xi \leq 0 }
\end{array} \rightarrow \left\{\begin{array}{c}
\xi \geq 1 \\
\xi \geq 2 / 5
\end{array} \rightarrow \xi \geq 1\right.\right.\right.
$$

Thus, event $A$ is realized if $\xi \in[1 ; 6]$ and therefore $P(A)=\frac{5}{6}$.

Answer: $P(A)=\frac{5}{6}$.