Black Jack Game Probability
Denote profit by $\pi$. Then
(a) $E[\pi] = (-1)\cdot P(\pi=-1)+(+1)\cdot P(\pi=1) = (-1)\cdot \frac{k}{5}+(+1)\cdot \frac{5-k}{5} = \frac{5-2k}{5}$
(b)$Var[\pi] = E[\pi^2] - \left(\frac{5-2k}{5}\right)^2 = \frac{k}{5} + \frac{5-k}{5} - \frac{(5-2k)^2}{25} = 1 - \frac{(5-2k)^2}{25} \to \min$
Solution is $k^* = 0$ or $k^* = 5$ (both yield $Var[\pi] = 0$)
(c) It is possible only if we have exactly 6 successes. Hence $$P(6 \text{ successes } ) = \binom{10}{6}\left(1-\frac{k}{5}\right)^6\left(\frac{k}{5}\right)^4$$