Pile of stones. Probability. Given a pile of stones. Every stone has a unique (different from others) weight. The $10$ easiest stones form $40\%$ of the total weight and $5$ most heavy-weight stones form $25\%$ of total weight. We randomly take one stone from the pile. What is the probability that the chosen stone will be the easiest (i.e belongs to the first $40\%$) ? I tried to use the law of total probability and and considered three hypotheses: $H_1$ --- the stone was from group of the easiest stones ( $40\%$); $H_2$ --- the stone was from group of the most heavy-weight stones stones ( $25\%$); $H_3$ --- the stone was from the rest part of the pile. ( $35\%$). But in this case I don't use the information: given $10$ and $5.$

The average weight of the 10 lightest stones is 4% of the total pile weight. The 5 heaviest stones have an average weight of 5 %. The remaining unknown number of stones must have an average between those values. So: $$ 4<\frac{35}{n}<5 \text{ with } n\in\mathbb{N} $$ With $n=8$ you get the probability: $$ p=\frac{10}{10 + 8 + 5}=\frac{10}{23} $$