I am reviewing for my upcoming test but I have difficulty in solving one problem because the way the define the variable X causing my confusion.I really need your help to build up a good solution for this problem.
Jack plays a game which assigns a number that is uniformly distributed between 0 and 5. If the game assigns a number less than or equal to k then he loses 1 dollar. On the other hand, if the game assigns a number larger than k, then he will gain 1 dollar.
(a) Find the expected profit of the game.
(b) What is the k value, which minimizes the variance of the profit obtained in this game?
(c) If you were to play this game 10 times, what is the probability that you gain 2 dollars?
Actually, I have thought that I will put X as the number of numbers which is less than or equal to k. Due to uniform distribution, I think I can build up a Probability distribution Function (PDF) can be expressed
1/6, if X in [0, k]
F(X) = 5/6, if X in (k, 5]
0, otherwise.
For this case, I can put Y according to X which shows the profit of Jack playing this game. However, I think the way I solve is wrong due to the expression of X.
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 $$P(6 \text{ successes } ) = \binom{10}{6}\left(1-\frac{k}{5}\right)^6\left(\frac{k}{5}\right)^4$$