A policeman inspects each passenger in the car that he stops during a roadblock. The number of passengers in a randomly stopped car is the random variable $X$. The probability distribution of $X$ is shown in the table below.

$x$	1	2	3	4	5
$\mathrm{P}(X = x)$	0.32	0.43	0.16	0.07	$k$

The policeman also checks the car boot if the number of passengers in the car is not more than 2.

1. Find the value of $k$ using the property that the sum of all probabilities is $1$

$\sum \mathrm{P}(X = x) = 1$
$0.32 + 0.43 + 0.16 + 0.07 + k = 1$
$0.98 + k = 1$
$k = 0.02$

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

$\mathrm{E}(X) = \sum x \mathrm{P}(X = x)$
$\mathrm{E}(X) = 1(0.32) + 2(0.43) + 3(0.16) + 4(0.07) + 5(0.02)$
$\mathrm{E}(X) = 0.32 + 0.86 + 0.48 + 0.28 + 0.10$
$\mathrm{E}(X) = 2.04$

1. Calculate the variance of $X$ using the value of $\mathrm{E}(X)$ from part (i)

From part (i), $k = 0.02$ and $\mathrm{E}(X) = 2.04$.

$\mathrm{E}(X^2) = 1^2(0.32) + 2^2(0.43) + 3^2(0.16) + 4^2(0.07) + 5^2(0.02)$
$\mathrm{E}(X^2) = 0.32 + 1.72 + 1.44 + 1.12 + 0.50 = 5.10$
$\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$
$\operatorname{Var}(X) = 5.10 - (2.04)^2 = 0.9384$

2. Calculate $\mathrm{E}(Y)$ and $\operatorname{Var}(Y)$ for the total passengers in 10 cars

Since $Y$ is the total number of passengers in 10 randomly stopped cars, it is the sum of 10 independent and identically distributed random variables, i.e., $Y = X_1 + X_2 + \dots + X_{10}$.

$\mathrm{E}(Y) = 10\mathrm{E}(X)$
$\mathrm{E}(Y) = 10(2.04) = 20.4$

$\operatorname{Var}(Y) = 10\operatorname{Var}(X)$
$\operatorname{Var}(Y) = 10(0.9384) = 9.384$

1. Find the probability that a single stopped car has its boot checked.

A car boot is checked if the number of passengers is not more than 2 ($X \le 2$).
$p = \mathrm{P}(X \le 2) = \mathrm{P}(X = 1) + \mathrm{P}(X = 2)$
$p = 0.32 + 0.43 = 0.75$

2. Define the binomial random variable and calculate the required probability.

Let $W$ be the number of car boots checked out of the 10 cars stopped.
$W \sim \mathrm{B}(10, 0.75)$
We want to find the probability that he checks more than 5 car boots:
$\mathrm{P}(W > 5) = 1 - \mathrm{P}(W \le 5)$
Using a graphing calculator:
$\mathrm{P}(W > 5) = 1 - 0.0781269 \dots$
$\mathrm{P}(W > 5) \approx 0.922 \text{ (to 3 s.f.)}$

1. Find the probability of checking a single car boot.

Let $p$ be the probability that the policeman checks a car boot.
The policeman checks the boot if $X \le 2$.
$p = \mathrm{P}(X \le 2) = \mathrm{P}(X = 1) + \mathrm{P}(X = 2)$
$p = 0.32 + 0.43 = 0.75$

2. Formulate the probability that exactly 4 boots are checked in the first 9 cars, and the 10th car boot is checked.

From part (ii), the total number of cars stopped at that roadblock is 10.
For the last (10th) car to be the 5th car boot checked, he must have checked exactly 4 car boots in the first 9 cars, followed by a boot check on the 10th car.
Let $W$ be the number of car boots checked in the first 9 cars. $W \sim \mathrm{B}(9, 0.75)$.
$\text{Probability} = \mathrm{P}(W = 4) \times p$
$\text{Probability} = \binom{9}{4}(0.75)^4(0.25)^5 \times 0.75$

3. Evaluate the final probability.

$\text{Probability} = 126 \times (0.75)^5 \times (0.25)^5$
$\text{Probability} = 0.029199\dots \approx 0.0292 \text{ (to 3 s.f.)}$