17. Three Forecasts (9th grade. 4 points). In Anchuria, days can be either clear or rainy, with the probability of a clear day being 0.74 regardless of the time of year. In the Committee on Climate and Forecasts of the Anchurian Parliament, there are two senators and one invited meteorologist. It is known that the proportion of correct forecasts of clear days for each of the three is the same as the proportion of correct forecasts of rainy days. However, the meteorologist's proportion is 50% higher than that of each senator.

One day, all three independently made a forecast for the approaching Day of Great Popular Celebration. Both senators stated that the day would be clear, while the meteorologist predicted rain. Which forecast should be trusted more?
Solution. Let event $G$ be "The VNL day will be clear", event $M_{1}$ "the first senator gave a forecast of 'clear'", event $M_{2}$ "the second senator gave a forecast of 'clear'", and event $S$ "the meteorologist gave a forecast of 'rain'". Let

$$
\mathrm{P}\left(M_{1} \mid G\right)=\mathrm{P}\left(M_{2} \mid G\right)=p
$$

Then from the condition, it follows that

$$
\begin{gathered}
\mathrm{P}(S \mid \bar{G})=\mathrm{P}(\bar{S} \mid G)=1.5 p \\
\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid G\right)=\mathrm{P}(S \mid G) \cdot \mathrm{P}\left(M_{1} \mid G\right) \cdot \mathrm{P}\left(M_{2} \mid G\right)=(1-1.5 p) p^{2}
\end{gathered}
$$

and

$$
\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid \bar{G}\right)=\mathrm{P}(S \mid \bar{G}) \cdot \mathrm{P}\left(M_{1} \mid \bar{G}\right) \cdot \mathrm{P}\left(M_{2} \mid \bar{G}\right)=1.5 p(1-p)^{2}
$$

To understand which forecast is better, we compare the conditional probabilities

$$
\mathrm{P}\left(\bar{G} \mid S \cap M_{1} \cap M_{2}\right)=\frac{\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid \bar{G}\right) \mathrm{P}(\bar{G})}{\mathrm{P}\left(S \cap M_{1} \cap M_{2}\right)}
$$

and

$$
\mathrm{P}\left(G \mid S \cap M_{1} \cap M_{2}\right)=\frac{\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid G\right) \mathrm{P}(G)}{\mathrm{P}\left(S \cap M_{1} \cap M_{2}\right)}
$$

If the first probability is higher, then the meteorologist's forecast is better; if the second is greater, then the senators are more likely to be correct.

Let $\mathrm{P}(G)=r$ (by condition $r=0.74$). Then

$$
\begin{aligned}
& \mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid \bar{G}\right) \mathrm{P}(\bar{G})=1.5 p(1-p)^{2}(1-r) \text { and } \\
& \quad \mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid G\right) \mathrm{P}(G)=(1-1.5 p) p^{2} r .
\end{aligned}
$$

Form the difference:

$$
\begin{gathered}
\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid \bar{G}\right) \mathrm{P}(\bar{G})-\mathrm{P}\left(S \cap M_{1} \cap M_{2} \mid G\right) \mathrm{P}(G)= \\
=1.5 p(1-p)^{2}(1-r)-(1-1.5 p) p^{2} r= \\
=1.5 p\left(1-2 p+p^{2}\right)-1.5 r p\left(1-2 p+p^{2}\right)-(1-1.5 p) p^{2} r= \\
=1.5 p-3 p^{2}+1.5 p^{3}-1.5 r p+3 r p^{2}-p^{2} r=p\left(1.5 p^{2}+(2 r-3) p+1.5(1-r)\right) .
\end{gathered}
$$

Substituting $r=0.74$, we get:

$$
p\left(1.5 p^{2}-1.52 p+0.39\right) .
$$

The discriminant of the quadratic trinomial in parentheses is

$$
1.52^{2}-4 \cdot 1.5 \cdot 0.39=-0.02960
$$

Therefore, in this case, it is better to rely on the forecast of one specialist rather than two politicians.

Answer: the meteorologist's forecast.