IB DP Mathematics • Exam Practice • 7 Marks

Statistics & Probability: Poisson Distribution & Conditional Probability

Curated examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

The number of meteorites entering Earth's atmosphere observed by an observatory follows a Poisson distribution with a mean rate of $2.4$ per hour. **(a)** [2 marks] Find the probability that exactly 3 meteorites are observed in a given one-hour window. **(b)** [2 marks] Find the probability that at least 5 meteorites are observed in a two-hour window. **(c)** [3 marks] Given that at least one meteorite is observed in a one-hour window, find the probability that more than 3 meteorites are observed.

Verified Solution & Marking Scheme

Part (a) [M1]
Let $X \sim \text{Po}(2.4)$. Formula: $P(X = 3) = \frac{e^{-2.4}(2.4)^3}{3!}$
Part (a) [A1]
$P(X = 3) \approx 0.209$
Part (b) [M1]
For two hours, $\lambda = 2 \times 2.4 = 4.8$. Let $Y \sim \text{Po}(4.8)$.
Part (b) [A1]
$P(Y \ge 5) = 1 - P(Y \le 4) = 1 - \text{poissoncdf}(4.8, 4) = 1 - 0.4763 = 0.524$
Part (c) [M1]
Conditional probability: $P(X > 3 \mid X \ge 1) = \frac{P(X > 3 \cap X \ge 1)}{P(X \ge 1)} = \frac{P(X \ge 4)}{1 - P(X = 0)}$
Part (c) [M1]
$P(X \ge 4) = 1 - \text{poissoncdf}(2.4, 3) = 1 - 0.7787 = 0.2213$. $P(X \ge 1) = 1 - e^{-2.4} = 1 - 0.0907 = 0.9093$.
Part (c) [A1]
$P(X > 3 \mid X \ge 1) = \frac{0.2213}{0.9093} \approx 0.243$
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