Problem Statement
Calls to an emergency dispatch service occur randomly and independently at an average rate of $3.5$ calls per $10$-minute interval.
(a) Find the probability of receiving at least 4 calls in a 10-minute interval. [2 marks]
(b) Find the probability of receiving exactly 8 calls in a 30-minute interval. [2 marks]
(c) Given that at least one call is received in a 10-minute interval, find the probability that at least 4 calls are received. [2 marks]
Verified Solution & Marking Scheme
Part (a): At least 4 calls in 10 mins
Let $X \sim \text{Po}(3.5)$.\n$$P(X \ge 4) = 1 - P(X \le 3) = 1 - \text{poissoncdf}(3.5, 3) = 1 - 0.53663 = 0.463$$
Part (b): 30-minute interval parameter
For 30 minutes, $\lambda = 3 \times 3.5 = 10.5$. Let $Y \sim \text{Po}(10.5)$.\n$$P(Y = 8) = \text{poissonpdf}(10.5, 8) = \frac{e^{-10.5}(10.5)^8}{8!} \approx 0.0988$$
Part (c): Conditional probability
$$P(X \ge 4 \mid X \ge 1) = \frac{P(X \ge 4 \cap X \ge 1)}{P(X \ge 1)} = \frac{P(X \ge 4)}{1 - P(X = 0)}$$\n$$P(X = 0) = e^{-3.5} = 0.030197 \implies P(X \ge 1) = 1 - 0.030197 = 0.9698$$\n$$P(X \ge 4 \mid X \ge 1) = \frac{0.46337}{0.9698} \approx 0.478$$