\(\displaystyle{E}{\left({X}\right)}={p}\cdot\cdot{\left({1}-{\left({1}-{p}\right)}^{{2}}\right)}+{\left({1}-{p}\cdot\right)}\cdot{\left({1}-{p}^{{2}}\right)}\)

\(\displaystyle={2}{p}{p}\cdot-{p}\cdot-{p}^{{2}}+{1}\)

We now have to analyse the function \(\displaystyle{p}\mapsto{E}{\left({X}\right)}\) and find its maximum on (0,1). We have that

\(\displaystyle{\frac{{{d}{E}{\left({X}\right)}}}{{{d}{p}}}}={2}{p}\cdot-{2}{p}={0}\)

\(\displaystyle\Leftrightarrow{p}={p}\cdot\)

So we have that the meteorologist maximezes it chances when \(\displaystyle{p}={p}\cdot\), which is according to our intuition.