11. Similar events. If a die is thrown, then the event $A=\{1,2,3\}$ consists of one of the faces 1, 2, or 3 appearing. Similarly, the event $B=\{1,2,4\}$ is favorable if the faces 1, 2, or 4 appear.

A die is thrown 10 times. It is known that the event $A$ occurred exactly 6 times.

a) (from 8th grade. 2 points) Find the probability that, under this condition, the event $B$ did not occur at all.

b) (from 9th grade. 3 points) Find the expected value of the random variable $X$ "The number of occurrences of event $B$".
Solution. a) The probability of the event "A happened six times" is $\frac{C_{10}^{6}}{2^{10}}$. The event "A happened six times and B never happened" will occur only if a three is rolled six times, and in the other cases, a 5 or 6 is rolled. The probability of this is $C_{10}^{6} \cdot\left(\frac{1}{6}\right)^{6} \cdot\left(\frac{2}{6}\right)^{4}=\frac{C_{10}^{6}}{3^{10} 2^{6}}$.

Therefore, the conditional probability of the event "B never happened" given that "A happened six times" is

$$
\frac{C_{10}^{6}}{3^{10} 2^{6}} \cdot \frac{2^{10}}{C_{10}^{6}}=\frac{2^{4}}{3^{10}} \approx 2.71 \cdot 10^{-4}
$$

b) First, consider only the 6 trials in which A occurred. Let $X_{1}$ be the number of these trials in which B occurred. If A occurred, then B occurs in two out of three cases (when 1 or 2 is rolled). Therefore, the probability of B given A is $\frac{2}{3}$. Hence, $\mathrm{E} X_{1}=\frac{2}{3} \cdot 6=4$.

Now consider the 4 trials in which A did not occur, i.e., 4, 5, or 6 points were rolled. Let $X_{2}$ be the number of these trials in which B occurred. If A did not occur, the probability of B is $\frac{1}{3}$ (only one outcome 4 out of three equally likely outcomes). Therefore, $\mathrm{E} X_{2}=\frac{1}{3} \cdot 4=\frac{4}{3}$.

If we denote $X$ as the total number of occurrences of B in ten rolls, then $X=X_{1}+X_{2}$, and therefore

$$
\mathrm{E} X=\mathrm{E} X_{1}+\mathrm{E} X_{2}=4+\frac{4}{3}=\frac{16}{3}
$$

Answer: $\frac{16}{3}$.