6. Vitya's Torments (from 7th grade, 2 points). Vitya has five math lessons in a week, one each day from Monday to Friday. Vitya knows that with a probability of $1 / 2$ the teacher will not check his homework during the school week, and with a probability of $1 / 2$ the teacher will check, but only during one of the math lessons, and it's impossible to guess when - on any lesson with equal chances.

By the end of the math lesson on Thursday, Vitya realized that the teacher had not yet checked his homework this week. What is the probability that the homework will be checked on Friday?
Answer: $\frac{1}{6}$.

Solution. Let's construct a tree for this random experiment (Fig.3).

![](https://cdn.mathpix.com/cropped/2024_05_06_142b35c881fc974fcd25g-04.jpg?height=463&width=1080&top_left_y=2056&top_left_x=568)

Fig. 3

Event $A$ "homework is not checked until Thursday" means that either there will be no check this week at all (the chain $S$ - "No check"), or the check will be on Friday (event $B$, represented by the chain $S-$ "There is a check" - Fri). We need to find the probability of event $B$ given $A$, which is the "proportion" of the chain $S-$ There is - Fri in event $A$:

$$
\mathrm{P}(B \mid A)=\frac{\mathrm{P}(S-\text { There is }- \text { Fri })}{\mathrm{P}(S-\text { There is }- \text { Fri })+\mathrm{P}(S-\text { No })}=\frac{1 / 2 \cdot 1 / 5}{1 / 2 \cdot 1 / 5+1 / 2}=\frac{1}{6} .
$$