12. Predictions (8th grade. 2 points). On Monday, Misha has five lessons, and on Tuesday - six. To determine which lessons unpredictable teachers will ask him about his homework, Misha tosses a coin 11 times - for the number of possible troubles. If it lands on heads, Misha assumes he will be asked on that lesson, if tails - he won't be asked. After the lessons on Tuesday, Misha noted that he guessed correctly 7 times. Find the probability that exactly 3 correct predictions referred to Monday.
Solution. The probability of guessing or not guessing a particular lesson is 0.5, regardless of the probability that the teacher will ask for Mishin's homework.

We are dealing with the conditional probability of the intersection of events $A_{1}$ "3 correct predictions on Monday" and $A_{2}$ "4 correct predictions on Tuesday" given $B$ "a total of 7 correct predictions":

$$
\mathrm{P}\left(A_{1} \cap A_{2} \mid B\right)=\frac{\mathrm{P}\left(A_{1} \cap A_{2} \cap B\right)}{\mathrm{P}(B)}
$$

Events $A_{1}$ and $A_{2}$ are independent, and their intersection implies $B$ (if $A_{1}$ and $A_{2}$ are true, then $B$ is true). Therefore,

$$
\mathrm{P}\left(A_{1} \cap A_{2} \mid B\right)=\frac{\mathrm{P}\left(A_{1} \cap A_{2}\right)}{\mathrm{P}(B)}=\frac{\mathrm{P}\left(A_{1}\right) \cdot \mathrm{P}\left(A_{2}\right)}{\mathrm{P}(B)}=\frac{C_{5}^{3} \frac{1}{2^{5}} \cdot C_{6}^{4} \frac{1}{2^{6}}}{C_{11}^{7} \frac{1}{2^{11}}}=\frac{C_{5}^{3} \cdot C_{6}^{4}}{C_{11}^{7}}=\frac{5}{11}
$$

Answer: $\frac{5}{11}$.