A roller-coaster ride has two separate safety systems to detect faults on the track and on the roller-coaster train itself. Over a long period of time, it is found that the average number of faults detected per day by the systems are 0.25 for the track and 0.15 for the train. Assume that the faults detected on the track are independent of those detected on the train.

1. State a necessary condition in the context of the question for a Poisson model to be valid.

For a Poisson distribution to accurately model the number of faults occurring, the events (faults) must occur singly, randomly, independently of each other, and at a constant average rate.

Stated in context, acceptable conditions include:

• The average rate of faults occurring per day remains constant from day to day.
• The occurrence of one fault does not affect the likelihood of another fault occurring (faults occur independently of one another).

1. Determine the distribution of the total number of faults in a 10-day period.

Let $X$ be the total number of faults detected by the two systems in 10 days.
The average number of total faults per day is $0.25 + 0.15 = 0.40$.
For a 10-day period, the mean number of faults is $\lambda = 0.40 \times 10 = 4$.
Given that the faults occur independently, the total number of faults follows a Poisson distribution, so $X \sim \text{Po}(4)$.

2. Calculate the probability of detecting at most 4 faults.

We require $\text{P}(X \le 4)$.
Using the Poisson cumulative distribution function (e.g., on a graphing calculator):
$\text{P}(X \le 4) \approx 0.62884$
$\text{P}(X \le 4) = 0.629 \text{ (to 3 s.f.)}$

1. Formulate an inequality for the probability of 0 faults occurring over $n$ days

Let $n$ be the number of days.
Since the total average number of faults per day is $0.25 + 0.15 = 0.4$, the number of faults in $n$ days follows a Poisson distribution with mean $0.4n$.
We require the probability of $0$ faults to be less than $0.05$:
$\mathrm{P}(Y = 0) < 0.05$
$\mathrm{e}^{-0.4n} < 0.05$

2. Solve the exponential inequality to find the smallest integer $n$

Taking the natural logarithm on both sides:
$-0.4n < \ln 0.05$
$n > \frac{\ln 0.05}{-0.4}$
$n > 7.4893$
Since $n$ must be an integer, the smallest number of days is $8$.

1. Identify the distributions for the track, train, and total faults in a 10-day period.

Let $X$, $Y$ and $T$ be the number of faults on the track, train, and in total respectively in 10 days.
$\lambda_X = 10 \times 0.25 = 2.5 \implies X \sim \text{Po}(2.5)$
$\lambda_Y = 10 \times 0.15 = 1.5 \implies Y \sim \text{Po}(1.5)$
$T = X + Y \sim \text{Po}(4)$

2. Set up the conditional probability.

We require $\mathrm{P}(X \ge 3 \mid T \le 4)$.
Using the formula for conditional probability:
$\mathrm{P}(X \ge 3 \mid T \le 4) = \frac{\mathrm{P}(X \ge 3 \text{ and } T \le 4)}{\mathrm{P}(T \le 4)}$

3. Enumerate the mutually exclusive cases satisfying both conditions.

Since $T = X + Y$ and $Y \ge 0$, the event $(X \ge 3 \text{ and } T \le 4)$ occurs under exactly three mutually exclusive scenarios:

• $X = 3, Y = 0$
• $X = 3, Y = 1$
• $X = 4, Y = 0$

4. Calculate the probabilities and evaluate the final conditional probability.

Since $X$ and $Y$ are independent:
$\begin{aligned} \mathrm{P}(X \ge 3 \text{ and } T \le 4) &= \mathrm{P}(X=3)\mathrm{P}(Y=0) + \mathrm{P}(X=3)\mathrm{P}(Y=1) + \mathrm{P}(X=4)\mathrm{P}(Y=0) \\ &= (0.21376)(0.22313) + (0.21376)(0.33470) + (0.13360)(0.22313) \\ &\approx 0.14905 \end{aligned}$
From the total distribution $T \sim \text{Po}(4)$, we have $\mathrm{P}(T \le 4) \approx 0.62884$ (often found in part ii).
$\mathrm{P}(X \ge 3 \mid T \le 4) = \frac{0.14905}{0.62884} \approx 0.23702 \approx 0.237 \text{ (to 3 s.f.)}$