Problem Statement
The number of customer support tickets received per hour by an IT firm follows a Poisson distribution with mean $\lambda = 4.5$.
(a) Find the probability that exactly 6 tickets are received in a given hour. [2 marks]
(b) Find the probability that at least 3 tickets are received in a given hour. [2 marks]
(c) Given that at least 3 tickets were received in a given hour, find the probability that fewer than 6 tickets were received. [2 marks]
Verified Solution & Marking Scheme
Part (a): Exactly 6 Tickets
$X \sim \text{Po}(4.5)$
$P(X = 6) = \frac{e^{-4.5} (4.5)^6}{6!} \approx 0.1281 \approx 0.128$
Part (b): At Least 3 Tickets
$P(X \ge 3) = 1 - P(X \le 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$
$P(X \le 2) = e^{-4.5} \left( 1 + 4.5 + \frac{4.5^2}{2} \right) = e^{-4.5} (1 + 4.5 + 10.125) = 15.625 e^{-4.5} \approx 0.1736$
$P(X \ge 3) = 1 - 0.1736 = 0.8264 \approx 0.826$
Part (c): Conditional Probability P(X < 6 | X ≥ 3)
$P(X < 6 \mid X \ge 3) = \frac{P(3 \le X \le 5)}{P(X \ge 3)}$
$P(3 \le X \le 5) = P(X \le 5) - P(X \le 2) = 0.7029 - 0.1736 = 0.5293$
$P(X < 6 \mid X \ge 3) = \frac{0.5293}{0.8264} \approx 0.6405 \approx 0.641$