Happie, the owner of a store selling novelty items, is organizing a publicity stunt for his store. Using a stock of ultra-rare Lubaba dolls that he acquired, he sets up a game where a player opens "mystery boxes" to try to find a Lubaba doll hidden inside one of the identical boxes used for the game. The rules of the game are:

• The player pays an initial $5 to start the game to open one of $n$ boxes.
• If the opened box is empty, the player can pay an additional $3 to open a second box.
• If the second box is empty, the player can pay $2.50 again to open a third box.
• If the third box is still empty, the player loses the game. The doll that is not won will then be donated away and the game is reset for the next player.

As the Lubaba dolls are considered rare collectors' items, it can be assumed that in every game, the players will keep opening the boxes until they win the doll and that they have the means to pay for the maximum of 3 allowable trys at opening the boxes. Let $X$ be the random variable denoting the number of empty boxes a player has opened in a game.

1. Apply the multiplication principle for dependent events to find the probability of opening 2 empty boxes followed by the box with the doll.

$\begin{aligned} P(X=2) &= P(\text{1st box empty}) \times P(\text{2nd box empty}) \times P(\text{3rd box has doll}) \\ &= \frac{n-1}{n} \times \frac{n-2}{n-1} \times \frac{1}{n-2} \\ &= \frac{1}{n} \end{aligned}$

1. Identify the possible values of $X$ and calculate $P(X=0)$ and $P(X=1)$.

Since a player can open a maximum of 3 boxes, the possible numbers of empty boxes opened ($X$) are $0, 1, 2,$ and $3$.

If $X=0$, the first box contains the doll:
$P(X=0) = \frac{1}{n}$

If $X=1$, the first box is empty and the second box contains the doll:
$P(X=1) = \frac{n-1}{n} \times \frac{1}{n-1} = \frac{1}{n}$

2. Determine $P(X=3)$ and present the probability distribution table.

Since the sum of all probabilities in a probability distribution must equal $1$, and we are given $P(X=2) = \frac{1}{n}$ from part (a):

$P(X=3) = 1 - P(X=0) - P(X=1) - P(X=2)$
$P(X=3) = 1 - \frac{1}{n} - \frac{1}{n} - \frac{1}{n} = \frac{n-3}{n}$

(Alternatively, calculating directly: $P(X=3) = \frac{n-1}{n} \times \frac{n-2}{n-1} \times \frac{n-3}{n-2} = \frac{n-3}{n}$)

The probability distribution of $X$ is:

1. Evaluate the probability distribution of $X$ for $n=10$.

From part (b), the probability distribution for $X$ is:
$P(X=0) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=1) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=2) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=3) = \frac{n-3}{n} = \frac{7}{10} = 0.7$

2. Calculate the expectation, variance, and standard deviation of $X$.

$E(X) = 0(0.1) + 1(0.1) + 2(0.1) + 3(0.7) = 2.4$
$E(X^2) = 0^2(0.1) + 1^2(0.1) + 2^2(0.1) + 3^2(0.7) = 6.8$
$\text{Var}(X) = E(X^2) - [E(X)]^2 = 6.8 - 2.4^2 = 1.04$
$\sigma = \sqrt{1.04} \approx 1.0198$

3. Evaluate $P(X \le \sigma)$.

Since $X$ only takes discrete integer values,
$P(X \le 1.0198) = P(X=0) + P(X=1)$
$= 0.1 + 0.1 = 0.2$

1. Determine the total revenue collected from a player for each possible value of $X$.

Let $R$ be the revenue collected per game.
If $X=0$, the player wins on the first box and pays $\char36{}5$.
If $X=1$, the player wins on the second box and pays $\char36{}5 + \char36{}3 = \char36{}8$.
If $X=2$, the player wins on the third box and pays $\char36{}5 + \char36{}3 + \char36{}2.50 = \char36{}10.50$.
If $X=3$, the player loses after opening $3$ empty boxes and pays $\char36{}5 + \char36{}3 + \char36{}2.50 = \char36{}10.50$.

2. Set up the equation for the expected revenue.

From the probability distribution of $X$ found in (b), $P(X=0)=\frac{1}{n}$, $P(X=1)=\frac{1}{n}$, $P(X=2)=\frac{1}{n}$, and $P(X=3)=\frac{n-3}{n}$.
The expected revenue is:
$\text{E}(R) = 5\left(\frac{1}{n}\right) + 8\left(\frac{1}{n}\right) + 10.5\left(\frac{1}{n}\right) + 10.5\left(\frac{n-3}{n}\right)$
$\text{E}(R) = \frac{5 + 8 + 10.5 + 10.5(n-3)}{n}$
$\text{E}(R) = \frac{23.5 + 10.5n - 31.5}{n} = \frac{10.5n - 8}{n}$

3. Equate the expected revenue to the cost of the doll and solve for $n$.

Since one doll is either won or donated away in every game, the expected cost per game is always $\char36{}10$.
For Happie to make no profit and incur no loss, the expected revenue must equal the cost per game ($\text{E}(R) = 10$).
$\frac{10.5n - 8}{n} = 10$
$10.5n - 8 = 10n$
$0.5n = 8$
$n = 16$