3. 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 types B and C require detailed answers. Since Ivan constantly skipped classes, his knowledge of mathematics is shallow. He answers tasks of type A by guessing. The first task of type B, Ivan solves with a probability of $\frac{1}{3}$. He cannot solve 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 version. (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.

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# Solution.

Event $A_{2 n}=\{$ Ivan Semenov receives $n$ points or more $\}$.

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

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

Obviously, the probability of correctly guessing one of 4 solutions is $\frac{1}{4}$. According to the Bernoulli distribution law $\mathrm{P}\left(A_{=i}\right)=C_{n}^{i} p^{i}(1-p)^{n-i}$, then

$$
\begin{aligned}
& \mathrm{P}\left(A_{4}\right)=\sum_{i=4}^{10} C_{10}^{i}\left(\frac{1}{4}\right)^{i}\left(\frac{3}{4}\right)^{10-i}=0,224124908 \\
& \mathrm{P}\left(A_{6}\right)=\sum_{i=6}^{10} C_{10}^{i}\left(\frac{1}{4}\right)^{i}\left(\frac{3}{4}\right)^{10-i}=0,019727707
\end{aligned}
$$

Therefore,

$$
\begin{gathered}
\mathrm{P}(\{\text { he gets more than } 5 \text { points }\})=0,224124908 \cdot \frac{1}{3}+0,019727707 \cdot \frac{2}{3}= \\
=0,087860107 \approx 0,088
\end{gathered}
$$

Answer: $\approx 0,088$.