55. Calves are delivered to a fattening complex from three farms. The number of calves from the first farm is twice as many as from the second, and the number from the second farm is three times as many as from the third. The first farm supplies $15 \%$ of the calves weighing more than 300 kg. The second and third farms supply $25 \%$ and $35 \%$ of the calves, respectively, whose live weight exceeds 300 kg. A randomly selected calf, upon arrival at the fattening complex, weighs $320 \mathrm{kg}$. What is the probability that it came from the third farm?
S o l u t i o n. The test consists of weighing a randomly selected calf from among those received at the fattening complex.

Consider the events:

$B_{1}, B_{2}, B_{3}$ - a calf from the 1st, 2nd, and 3rd farm, respectively, has been selected;

$A$ - a randomly selected calf has a live weight exceeding $300 \mathrm{kr}$.

Let's find the unconditional probabilities of the hypotheses - events $B_{1}, B_{2}$, and $B_{3}$ before the test. Assuming that $x \%$ of the calves come from the 3rd farm, we get that $3 x \%$ and $6 x \%$ of the calves come from the 2nd and 3rd farms, respectively. The total number of calves from these three farms is $100 \%$. Thus, we have: $10 x=100, x=10$. Therefore, the number of calves from the 1st, 2nd, and 3rd farms is $60 \%$, $30 \%$, and $10 \%$ of the total, respectively. Taking this into account, we find the probabilities of the hypotheses before the test:

$$
P\left(B_{1}\right)=0.6 ; P\left(B_{2}\right)=0.3 ; P\left(B_{3}\right)=0.1
$$

From the 1st farm, $15 \%$ of the calves have a live weight exceeding $300 \mathrm{kr}$, so $P_{B_{1}}(A)=0.15$. Similarly, we find: $P_{B_{2}}(A)=0.25$, $P_{B_{3}}(A)=0.35$.

Event $A$ can occur together with event $B_{1}$, or with event $B_{2}$, or with event $B_{3}$.

Suppose event $A$ has occurred. Re-evaluate the probability of the third hypothesis - event $B_{3}$.

Using formula (17) for $n=3$, we find:

$$
\begin{aligned}
& P_{A}\left(B_{3}\right)=\frac{P\left(B_{3}\right) P_{B_{3}}(A)}{P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)+P\left(B_{3}\right) P_{B_{3}}(A)}= \\
& =\frac{0.1 \cdot 0.35}{0.6 \cdot 0.15+0.3 \cdot 0.25+0.1 \cdot 0.35}=0.175
\end{aligned}
$$