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 tries at opening the boxes. Let $X$ be the random variable denoting the number of empty boxes a player has opened in a game.

1. Identify the sequence of events that results in exactly two empty boxes being opened.

For $X=2$, the player opens exactly 2 empty boxes before the game ends. Since players keep opening boxes until they win the doll or reach the maximum of 3 tries, having exactly 2 empty boxes means the first two boxes were empty and the third box contained the doll.
$\text{P}(X=2) = \text{P}(\text{1st empty}) \times \text{P}(\text{2nd empty}) \times \text{P}(\text{3rd is doll})$

2. Calculate the probability using without-replacement multiplication.

There is 1 doll and $n-1$ empty boxes initially.
$\text{P}(X=2) = \left(\frac{n-1}{n}\right) \times \left(\frac{n-2}{n-1}\right) \times \left(\frac{1}{n-2}\right)$
Cancelling out the common factors $(n-1)$ and $(n-2)$:
$\text{P}(X=2) = \frac{1}{n}\ \text{ (shown)}$

1. Determine the possible values of $X$ and calculate $\text{P}(X=0)$ and $\text{P}(X=1)$.

The possible values of $X$ (number of empty boxes opened) are $0, 1, 2, 3$.
$\text{P}(X = 0) = \text{P}(\text{doll in 1st box}) = \frac{1}{n}$
$\text{P}(X = 1) = \text{P}(\text{1st empty}) \times \text{P}(\text{2nd doll}) = \frac{n-1}{n} \times \frac{1}{n-1} = \frac{1}{n}$

2. Calculate $\text{P}(X=3)$ using the property that the sum of all probabilities is 1.

Since the game ends after a maximum of 3 tries, $\sum \text{P}(X=x) = 1$.
Using $\text{P}(X=2) = \frac{1}{n}$ from part (a):
$\text{P}(X = 3) = 1 - \text{P}(X=0) - \text{P}(X=1) - \text{P}(X=2)$
$\text{P}(X = 3) = 1 - \frac{1}{n} - \frac{1}{n} - \frac{1}{n} = \frac{n-3}{n}$

3. Present the probability distribution in a table.

1. Substitute $n = 10$ into the probability distribution of $X$.

From part (b), the probability distribution of $X$ is:
$\text{P}(X = 0) = \frac{1}{n}$
$\text{P}(X = 1) = \frac{1}{n}$
$\text{P}(X = 2) = \frac{1}{n}$
$\text{P}(X = 3) = \frac{n-3}{n}$

When $n = 10$:
$\text{P}(X = 0) = \frac{1}{10}$
$\text{P}(X = 1) = \frac{1}{10}$
$\text{P}(X = 2) = \frac{1}{10}$
$\text{P}(X = 3) = \frac{7}{10}$

2. Calculate the expected value $\text{E}(X)$ and variance $\text{Var}(X)$.

$\text{E}(X) = \sum x \,\text{P}(X=x)$
$\text{E}(X) = 0\left(\frac{1}{10}\right) + 1\left(\frac{1}{10}\right) + 2\left(\frac{1}{10}\right) + 3\left(\frac{7}{10}\right) = \frac{0 + 1 + 2 + 21}{10} = 2.4$

$\text{E}(X^2) = \sum x^2 \,\text{P}(X=x)$
$\text{E}(X^2) = 0^2\left(\frac{1}{10}\right) + 1^2\left(\frac{1}{10}\right) + 2^2\left(\frac{1}{10}\right) + 3^2\left(\frac{7}{10}\right) = \frac{0 + 1 + 4 + 63}{10} = 6.8$

$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$
$\text{Var}(X) = 6.8 - (2.4)^2 = 6.8 - 5.76 = 1.04$

3. Find the standard deviation $\sigma$ and evaluate $\text{P}(X \le \sigma)$.

$\sigma = \sqrt{1.04} \approx 1.0198$

Since $X$ takes non-negative integer values:
$\text{P}(X \le \sigma) = \text{P}(X \le 1.0198)$
$\text{P}(X \le \sigma) = \text{P}(X = 0) + \text{P}(X = 1)$
$\text{P}(X \le \sigma) = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = 0.2$

1. Determine the probabilities and the total revenue collected for each outcome.

From part (b), the probability distribution for the number of empty boxes opened ($X$) is:
$\text{P}(X=0) = \frac{1}{n}$
$\text{P}(X=1) = \frac{1}{n}$
$\text{P}(X=2) = \frac{1}{n}$
$\text{P}(X=3) = \frac{n-3}{n}$

Let $R$ be the revenue collected per game. According to the rules, the revenue for each outcome is:
If $X=0$ (finds doll in 1st box): $R = 5$
If $X=1$ (finds doll in 2nd box): $R = 5 + 3 = 8$
If $X=2$ (finds doll in 3rd box): $R = 5 + 3 + 2.50 = 10.50$
If $X=3$ (3 boxes empty, game lost): $R = 5 + 3 + 2.50 = 10.50$

2. Formulate an equation using the expected revenue and the expected cost.

For Happie to make no profit and incur no loss, the expected revenue per game must equal the cost of the doll used up per game (which is given away or donated either way).
$\text{E}(R) = \text{Cost per game}$
$5\left(\frac{1}{n}\right) + 8\left(\frac{1}{n}\right) + 10.50\left(\frac{1}{n}\right) + 10.50\left(\frac{n-3}{n}\right) = 10$

3. Solve for the number of boxes $n$.

$\frac{5 + 8 + 10.50 + 10.50(n-3)}{n} = 10$
$\frac{23.5 + 10.50n - 31.50}{n} = 10$
$\frac{10.50n - 8}{n} = 10$
$10.50n - 8 = 10n$
$0.50n = 8$
$n = 16$