8. Reviews (from 7th grade. 1 point). Angry reviews about the work of an online store are left by $80 \%$ of dissatisfied customers (those who were poorly served in the store). Only $15 \%$ of satisfied customers leave positive reviews. A certain online store has received 60 angry and 20 positive reviews. Using this statistics, estimate the probability that the next customer will be satisfied with the service in this online store[^0]

![](https://cdn.mathpix.com/cropped/2024_05_06_cb774fbc1caae3aa402cg-03.jpg?height=574&width=654&top_left_y=287&top_left_x=241)
Solution. Let $p$ be the probability that a customer is served well, and $q=1-p$ be the probability that they are served poorly. Then the probability that a customer leaves a good review is $0.15 p$, and the probability that there will be a bad review is $0.8(1-p)$. Then

$$
\frac{0.15 p}{0.8(1-p)} \approx \frac{1}{3}, \text{ hence } 25 p \approx 16, p \approx 0.64
$$

Answer: approximately 0.64.