Event $A_{n}=\{$ Ivan Semenov receives $n$ points or more for tasks of type $A \}$.

Event $B=\{$ He solves a problem of type $B\}$.

Then $P(\{$ Ivan received more than 5 points $\})=P\left(A_{4}\right) P(B)+P\left(A_{6}\right) P(\bar{B})$.

The probability of correctly guessing one of 4 solutions to a specific problem is $1 / 4$. According to the Bernoulli distribution law, $P\left(A_{4}\right)=\sum_{i=4}^{10} C_{10}^{i}\left(\frac{1}{4}\right)^{i}\left(\frac{3}{4}\right)^{10-i}, P\left(A_{6}\right)=\sum_{i=6}^{10} C_{10}^{i}\left(\frac{1}{4}\right)^{i}\left(\frac{3}{4}\right)^{10-i}$.

Therefore, $P(\{$ Ivan received more than 5 points $\})=1 / 3 P\left(A_{4}\right)+2 / 3 P\left(A_{6}\right)=$

$$
\frac{1}{3 \cdot 4^{10}}\left(C_{10}^{4} 3^{6}+C_{10}^{5} 3^{5}+3\left(C_{10}^{6} 3^{4}+C_{10}^{7} 3^{3}+C_{10}^{2} 3^{2}+10 \cdot 3+1\right)\right) \approx 0.088
$$

## Answer

$\approx 0.088$.
Ivan Semenov is taking a Unified State Exam (USE) in mathematics. The exam consists of three types of tasks: $A, B$, and $C$. For each task of type $A$, there are four answer options, only one of which is correct. There are 10 such tasks in total. Tasks of type $B$ and $C$ require detailed answers. Since Ivan constantly skipped classes, his knowledge of mathematics is shallow. He completes tasks of type $A$ by randomly selecting answers. The first task of type $B$ Ivan solves with a probability of $1 / 2$. He cannot do anything else. For a correct answer to one task of type $A$, 1 point is awarded, and for a task of type $B$, 2 points. What is the probability that Ivan will score more than 5 points?

Take the tasks of type $A$ from the 2008 USE practice test. (http://ege.edu.ru/demo/math.zip) and conduct the experiment of randomly selecting answers 10 times. Compare the result with the theoretical one (for 5 correct answers). Ensure that the results do not differ significantly.