Pat is invited to a game show. In the game, she is presented with a box with $k$ balls with either 1, 2, 3 or 4 printed on each of the balls. The probability of drawing a ball with number $i$ printed on it is $\frac{i}{k}$, where $i = 1, 2, 3, 4$. Two balls are drawn one after another with replacement. The prize money that Pat gets is $W, where $W$ is the mean of the two numbers printed on the balls drawn.

1. Set the sum of the probabilities of all possible outcomes to 1

$\frac{1}{k} + \frac{2}{k} + \frac{3}{k} + \frac{4}{k} = 1$

2. Solve the equation for $k$

$\frac{1+2+3+4}{k} = 1$
$\frac{10}{k} = 1$
$k = 10$

1. Identify the possible values of $W$ by considering all combinations of the two drawn numbers.

Let $X_1$ and $X_2$ be the numbers drawn on the first and second draws, where $X_1, X_2 \in \{1, 2, 3, 4\}$.
Since $W$ is the mean of the two numbers, $W = \frac{X_1 + X_2}{2}$.
The possible sums of $X_1 + X_2$ are $2, 3, 4, 5, 6, 7, 8$, so the possible values of $W$ are $1, 1.5, 2, 2.5, 3, 3.5, 4$.

2. Calculate the probability for each value of $W$ using the independence of the two draws.

Since the balls are drawn with replacement, the events are independent. Thus, $\mathrm{P}(X_1=a, X_2=b) = \mathrm{P}(X_1=a) \times \mathrm{P}(X_2=b) = \frac{a}{10} \times \frac{b}{10} = \frac{ab}{100}$.

Calculating for each $W$:
$\mathrm{P}(W=1) = \mathrm{P}(1,1) = \frac{1 \times 1}{100} = \frac{1}{100}$

$\mathrm{P}(W=1.5) = \mathrm{P}(1,2) + \mathrm{P}(2,1) = \frac{1 \times 2}{100} + \frac{2 \times 1}{100} = \frac{4}{100} = \frac{1}{25}$

$\mathrm{P}(W=2) = \mathrm{P}(1,3) + \mathrm{P}(3,1) + \mathrm{P}(2,2) = \frac{3}{100} + \frac{3}{100} + \frac{4}{100} = \frac{10}{100} = \frac{1}{10}$

$\mathrm{P}(W=2.5) = \mathrm{P}(1,4) + \mathrm{P}(4,1) + \mathrm{P}(2,3) + \mathrm{P}(3,2) = \frac{4}{100} + \frac{4}{100} + \frac{6}{100} + \frac{6}{100} = \frac{20}{100} = \frac{1}{5}$

$\mathrm{P}(W=3) = \mathrm{P}(2,4) + \mathrm{P}(4,2) + \mathrm{P}(3,3) = \frac{8}{100} + \frac{8}{100} + \frac{9}{100} = \frac{25}{100} = \frac{1}{4}$

$\mathrm{P}(W=3.5) = \mathrm{P}(3,4) + \mathrm{P}(4,3) = \frac{12}{100} + \frac{12}{100} = \frac{24}{100} = \frac{6}{25}$

$\mathrm{P}(W=4) = \mathrm{P}(4,4) = \frac{16}{100} = \frac{4}{25}$

3. Present the probability distribution of $W$ in a table format.

(Self-check: $\frac{1}{100} + \frac{4}{100} + \frac{10}{100} + \frac{20}{100} + \frac{25}{100} + \frac{24}{100} + \frac{16}{100} = \frac{100}{100} = 1$)

1. Calculate the population mean (the mean of all numbers printed on the balls in the box).

Let $\theta$ be the mean of all numbers printed on the balls in the box. Since there are $k=10$ balls (from part i) with numbers 1, 2, 3, and 4 having frequencies 1, 2, 3, and 4 respectively:
$\theta = \frac{1(1) + 2(2) + 3(3) + 4(4)}{10} = \frac{30}{10} = 3$

2. Calculate the expected value of $W$.

$W$ is the mean of two independent draws $X_1$ and $X_2$, so $W = \frac{X_1 + X_2}{2}$.
$E(W) = E\left(\frac{X_1 + X_2}{2}\right) = \frac{E(X_1) + E(X_2)}{2}$
Since $E(X_1) = E(X_2) = \theta = 3$,
$E(W) = \frac{3 + 3}{2} = 3$
(Note: Alternatively, using the probability distribution of $W$ from part (ii): $E(W) = \sum w P(W=w) = 1(0.01) + 1.5(0.04) + 2(0.10) + 2.5(0.20) + 3(0.25) + 3.5(0.24) + 4(0.16) = 3$)

3. Conclude that $W$ is an unbiased estimator.

Since $E(W) = \theta = 3$, $W$ satisfies the given definition of an unbiased estimator of the population mean.

1. Find the probability that Pat earns more than $2.50 in a single game.

Pat earns more than $\char36{}2.50$ if the mean of the two numbers drawn is strictly greater than $2.5$, which means their sum must be $6, 7,$ or $8$. From earlier parts, the probability of drawing number $i$ is $\frac{i}{10}$.
The probabilities for these outcomes are:

• Sum $= 6$: $(2,4), (4,2), (3,3) \Rightarrow P(\text{sum } 6) = 2(0.2)(0.4) + (0.3)^2 = 0.25$
• Sum $= 7$: $(3,4), (4,3) \Rightarrow P(\text{sum } 7) = 2(0.3)(0.4) = 0.24$
• Sum $= 8$: $(4,4) \Rightarrow P(\text{sum } 8) = (0.4)^2 = 0.16$

$P(W > 2.50) = 0.25 + 0.24 + 0.16 = 0.65$

2. Define the binomial distribution for the 5 games.

Let $Y$ be the number of games out of $5$ in which Pat earns more than $\char36{}2.50$.
Since the games are independent and the probability of success is constant, $Y$ follows a binomial distribution:
$Y \sim \text{B}(5, 0.65)$

3. Calculate the probability that Pat proceeds to the second round.

Pat proceeds if she earns more than $\char36{}2.50$ in at least $3$ games, so we need $P(Y \ge 3)$.
$P(Y \ge 3) = P(Y=3) + P(Y=4) + P(Y=5)$
$= \binom{5}{3}(0.65)^3(0.35)^2 + \binom{5}{4}(0.65)^4(0.35)^1 + \binom{5}{5}(0.65)^5(0.35)^0$
$= 10(0.274625)(0.1225) + 5(0.17850625)(0.35) + 1(0.1160290625)$
$= 0.336415625 + 0.3123859375 + 0.1160290625$
$= 0.764830625$
$\approx 0.765 \text{ (to 3 s.f.)}$

1. Formulate the expected amount earned in the second round in terms of $m$.

The probability of drawing a black ball is $\frac{2}{m}$ and the probability of drawing a white ball is $\frac{m-2}{m}$.

$\text{Expected amount} = \left(\frac{2}{m}\right)(3m) + \left(\frac{m-2}{m}\right)\left(-\frac{m}{2}\right)$

2. Set up and solve the inequality for a non-negative expected amount.

$\left(\frac{2}{m}\right)(3m) - \left(\frac{m-2}{m}\right)\left(\frac{m}{2}\right) \ge 0$
$6 - \frac{m-2}{2} \ge 0$
$12 - m + 2 \ge 0$
$14 - m \ge 0$
$m \le 14$

3. Determine the lower bound for $m$ based on the physical constraints to find the full range.

Since there is at least one white ball in the box, the number of white balls must be at least $1$:
$m - 2 \ge 1$
$m \ge 3$

Combining the two inequalities, we get $3 \le m \le 14$. Since $m$ represents a number of balls, it must be an integer.