In a game, a player selects one ball at random from a bag which contains four balls numbered '1', '1', '2' and '3'. The player notes the number on the selected ball, and then tosses that number of fair coins. The number of heads obtained is denoted by $X$ and has a probability distribution as follows:

$x$	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:
$\mathrm{P}(\bar{X} > 1) \approx 0.12878$
$\mathrm{P}(\bar{X} > 1) = 0.129 \text{ (to 3 s.f.)}$