A biased die in the form of a tetrahedron has its four faces labelled 1 to 4, with one number printed on each face. The die is tossed and $X$ is the random variable representing the number on the face on which the die lands. The probability distribution of $X$ is shown in the table below.

$x$	1	2	3	4
$\mathrm{P}(X = x)$	$p$	$q$	$q$	$p$

A game is played by a player tossing this die once. If the die lands on the face with the number 4 printed on it, the player wins, otherwise the player loses. The random variable $Y$ denotes the number of wins out of 50 games the player plays.

1. Determine the value of $\mathrm{E}(X)$ using the sum of probabilities or symmetry

Since the sum of all probabilities is $1$:
$2p + 2q = 1 \implies p + q = 0.5$

Using the formula for expectation:
$\mathrm{E}(X) = 1(p) + 2(q) + 3(q) + 4(p)$
$\mathrm{E}(X) = 5p + 5q = 5(p + q)$
$\mathrm{E}(X) = 5(0.5) = 2.5$

(Alternatively, by observing that the probability distribution is perfectly symmetrical about $2.5$, one can directly state $\mathrm{E}(X) = 2.5$.)

1. Establish the first equation using the sum of probabilities.

$\sum \mathrm{P}(X=x) = 1$
$2p + 2q = 1$
$p + q = 0.5 \implies q = 0.5 - p$

2. Express $\mathrm{E}(X^2)$ in terms of $p$ and $q$.

$\mathrm{E}(X^2) = 1^2(p) + 2^2(q) + 3^2(q) + 4^2(p)$
$= p + 4q + 9q + 16p$
$= 17p + 13q$

3. Form a second equation using the given variance and the value of $\mathrm{E}(X)$ from part (a).

From part (a), by symmetry, $\mathrm{E}(X) = 2.5$.
$\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$
$1.65 = (17p + 13q) - (2.5)^2$
$1.65 = 17p + 13q - 6.25$
$17p + 13q = 7.9$

4. Solve the simultaneous equations to find $p$ and $q$.

Substitute $q = 0.5 - p$ into the variance equation:
$17p + 13(0.5 - p) = 7.9$
$17p + 6.5 - 13p = 7.9$
$4p = 1.4$
$p = 0.35$

Substitute $p = 0.35$ back to find $q$:
$q = 0.5 - 0.35 = 0.15$

1. State the assumption of independence between trials

The outcome of each game (whether the player wins or loses) must be independent of all other games.

2. State the assumption of a constant probability of success

The probability of winning (getting a 4) must remain constant for every game played.

1. Find the value of $p$ using the probability distribution and conditions from earlier parts.

From the probability table, the sum of probabilities is $1$:
$2p + 2q = 1 \implies p + q = 0.5$
The expected value is:
$\mathrm{E}(X) = 1(p) + 2(q) + 3(q) + 4(p) = 5p + 5q = 5(0.5) = 2.5$
Using the given $\operatorname{Var}(X) = 1.65$:
$\mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 1.65$
$[1^2(p) + 2^2(q) + 3^2(q) + 4^2(p)] - 2.5^2 = 1.65$
$17p + 13q = 1.65 + 6.25 = 7.9$
Substituting $q = 0.5 - p$:
$17p + 13(0.5 - p) = 7.9$
$4p + 6.5 = 7.9 \implies 4p = 1.4 \implies p = 0.35$

2. Define the distribution of $Y$ and find the expected number of games won.

The probability of winning a game (landing on 4) is $p = 0.35$.
Given $50$ games, $Y \sim \mathrm{B}(50, 0.35)$.

The expected number of games won is:
$\mathrm{E}(Y) = np = 50 \times 0.35 = 17.5$

3. Calculate the probability that the player wins more than the expected number of games.

We require $\mathrm{P}(Y > 17.5)$. Since $Y$ is a discrete integer count:
$\mathrm{P}(Y > 17.5) = \mathrm{P}(Y \ge 18)$
$\mathrm{P}(Y \ge 18) = 1 - \mathrm{P}(Y \le 17)$
Using a graphing calculator:
$\mathrm{P}(Y \ge 18) \approx 1 - 0.499066 = 0.500934$
$\mathrm{P}(Y \ge 18) \approx 0.501 \text{ (to 3 s.f.)}$