Let
P(A) = x, P(B) = 1-x ($\because$ A and B are mutually exclusive and A $\cup$ B = S)
We want to find max value of
f(x) = P(A)*P(B) = x*(1-x) = x – $x^{2}$
we find critical points i.e $f^{'}(x)$ = 0
$f^{'}(x)$ = 1-2x = 0
$\therefore$ x = ½ , this x can be minima or maxima, so we will find second derivative and if $f^{''}(x)$ < 0, then x will give max value for $f(x)$
$f^{''}(x)$ = -2 < 0
$\therefore$ at x = ½ we get max value
=x(1-x) = ½ * ½ = ¼ = 0.25