15. Mysterious Animal (from 9th grade. 2 points). Exactly half of the population of the Island of Misfortune are rabbits, and all the others are hares. If a resident of the Island of Misfortune asserts something, they always sincerely believe what they say. However, hares are honestly mistaken on average in one out of every four cases, while rabbits are honestly mistaken on average in one out of every three cases. One day, an animal emerged in the center of the island and shouted: “I am not a hare!” It then thought and sadly said: “I am not a rabbit.” What is the probability that it is still a hare?

![](https://cdn.mathpix.com/cropped/2024_05_06_80571d39ca980f497d63g-11.jpg?height=368&width=349&top_left_y=244&top_left_x=1573)
Solution. Let's consider the events $A$ "The animal is a hare", $B$ "The animal claimed it is not a hare", $C$ "The animal claimed it is not a rabbit".

We need to find the conditional probability $\mathrm{P}(A \mid B \cap C)$:

$$
\begin{aligned}
\mathrm{P}(A \mid B \cap C) & =\frac{\mathrm{P}(A \cap B \cap C)}{\mathrm{P}(B \cap C)}=\frac{\mathrm{P}(A \cap B \cap C)}{\mathrm{P}(A \cap B \cap C)+\mathrm{P}(\bar{A} \cap B \cap C)}= \\
& =\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{3}}=\frac{\frac{3}{16}}{\frac{3}{16}+\frac{2}{9}}=\frac{27}{59} \approx 0.458
\end{aligned}
$$

Answer: approximately 0.458.