在一个游戏中，玩家从一个装有四个分别标有“1”、“1”、“2”和“3”的球的袋子中随机选择一个球。玩家记录所选球上的数字，然后投掷该数字个均匀对称的硬币。得到的正面朝上的硬币个数记为 $X$，其概率分布如下：

$x$	0	1	2	3
$\mathrm{P}(X = x)$	$a$	$\frac{15}{32}$	$b$	$\frac{1}{32}$

1. Calculate the probability of obtaining 0 heads to find $a$.

Let the number on the selected ball be $B$.

2. Use the property that the sum of all probabilities in a distribution is 1 to find $b$.

1. Determine the required probabilities $a$ and $b$ from part (i) to complete the probability distribution.

To calculate the expectation, we first need the value of $b$ (and we can find $a$ for completeness) from part (i).
$b = \mathrm{P}(X=2)$
This can occur if ball '2' is selected and 2 heads are obtained, or if ball '3' is selected and 2 heads are obtained:
$b = \mathrm{P}(\text{ball } 2) \cdot \mathrm{P}(2\text{ heads from } 2 \text{ coins}) + \mathrm{P}(\text{ball } 3) \cdot \mathrm{P}(2\text{ heads from } 3 \text{ coins})$
$b = \left(\frac{1}{4}\right)\left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)\left[\binom{3}{2}\left(\frac{1}{2}\right)^3\right]$
$b = \frac{1}{16} + \frac{3}{32} = \frac{5}{32}$
Since the probabilities must sum to 1:
$a + \frac{15}{32} + \frac{5}{32} + \frac{1}{32} = 1 \implies a = \frac{11}{32}$

2. Calculate the expected value $\mathrm{E}(X)$.

$\mathrm{E}(X) = \sum x \,\mathrm{P}(X=x)$
$\mathrm{E}(X) = 0\left(\frac{11}{32}\right) + 1\left(\frac{15}{32}\right) + 2\left(\frac{5}{32}\right) + 3\left(\frac{1}{32}\right)$
$\mathrm{E}(X) = \frac{0 + 15 + 10 + 3}{32} = \frac{28}{32}$
$\mathrm{E}(X) = \frac{7}{8}$

3. Calculate $\mathrm{E}(X^2)$ as an intermediate step for the variance.

$\mathrm{E}(X^2) = \sum x^2 \,\mathrm{P}(X=x)$
$\mathrm{E}(X^2) = 0^2\left(\frac{11}{32}\right) + 1^2\left(\frac{15}{32}\right) + 2^2\left(\frac{5}{32}\right) + 3^2\left(\frac{1}{32}\right)$
$\mathrm{E}(X^2) = \frac{0 + 15 + 20 + 9}{32} = \frac{44}{32}$
$\mathrm{E}(X^2) = \frac{11}{8}$

4. Calculate $\operatorname{Var}(X)$ and show that it matches the given value.

$\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$
$\operatorname{Var}(X) = \frac{11}{8} - \left(\frac{7}{8}\right)^2$
$\operatorname{Var}(X) = \frac{88}{64} - \frac{49}{64}$
$\operatorname{Var}(X) = \frac{39}{64} \quad \text{(shown)}$

1. Calculate the expected value $\mathrm{E}(X)$ required from part (ii).

To find $\mathrm{E}(X)$, we first determine $b = \mathrm{P}(X=2)$ from part (i). A player can only get 2 heads if they select ball '2' (probability $\frac{1}{4}$) or ball '3' (probability $\frac{1}{4}$).
$b = \frac{1}{4}\mathrm{P}(2 \text{ heads from } 2 \text{ coins}) + \frac{1}{4}\mathrm{P}(2 \text{ heads from } 3 \text{ coins})$
$b = \frac{1}{4}\left(\frac{1}{4}\right) + \frac{1}{4}\left(\binom{3}{2}\frac{1}{8}\right) = \frac{1}{16} + \frac{3}{32} = \frac{5}{32}$
Using the probability distribution for part (ii):
$\mathrm{E}(X) = 0(a) + 1\left(\frac{15}{32}\right) + 2\left(\frac{5}{32}\right) + 3\left(\frac{1}{32}\right) = \frac{15}{32} + \frac{10}{32} + \frac{3}{32} = \frac{7}{8}$
We are also given $\operatorname{Var}(X) = \frac{39}{64}$ from part (ii).

2. Identify the distribution of the sample mean $\bar{X}$ using the Central Limit Theorem.

Let $\bar{X}$ be the average number of heads obtained per game.
Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
$\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right)$
$\bar{X} \sim \mathrm{N}\left(\frac{7}{8}, \frac{39/64}{50}\right)$
$\bar{X} \sim \mathrm{N}\left(\frac{7}{8}, \frac{39}{3200}\right)$

3. Calculate the probability that the average number of heads exceeds 1.

We require $\mathrm{P}(\bar{X} > 1)$.
Using a graphing calculator:
$\mathrm{P}(\bar{X} > 1) \approx 0.12878$
$\mathrm{P}(\bar{X} > 1) = 0.129 \text{ (to 3 s.f.)}$