AS & A Level · Practice questions · The Poisson distribution

Count the random events.

Six original Cambridge-style Paper 6 questions on the Poisson distribution, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Define \(X\), fix \(\lambda\) for the interval, then calculate.

Define: "Let \(X\) be the number of ... in ..." and write \(X \sim \text{Po}(\lambda)\).
Scale: multiply the rate by the length of the interval in the question.
Boundaries: write the event as an inequality (\(X \ge 3\), \(X \le 2\)) before calculating.
Complements: \(\text{P}(X \ge k) = 1 - \text{P}(X \le k - 1)\).
Context: state conditions and conclusions in terms of the situation.
Accuracy: keep full calculator values; give answers to 3 s.f.

Recall

[2 marks]

Customers arrive at a bank. State two conditions needed for the number of customers arriving in a 5-minute period to be modelled by a Poisson distribution, in context.

Customers arrive independently of each other (not in groups). ✓
Customers arrive at a constant mean rate during the period considered (and singly, at random). ✓

Calculation

[4 marks]

\(X \sim \text{Po}(3.5)\). Find (a) \(\text{P}(X = 4)\) (b) \(\text{P}(2 \le X \le 4)\).

(a) \(e^{-3.5}\dfrac{3.5^{4}}{24} = 0.030\,197 \times 6.2526 = 0.189\). ✓
(b) \(\text{P}(2) + \text{P}(3) + \text{P}(4) = e^{-3.5}\left(\dfrac{3.5^{2}}{2} + \dfrac{3.5^{3}}{6} + \dfrac{3.5^{4}}{24}\right)\) ✓
\(= 0.030\,197 \times (6.125 + 7.1458 + 6.2526) = 0.030\,197 \times 19.523 = 0.590\). ✓

Calculation

[3 marks]

Emails arrive in an inbox at random at a mean rate of 9 per hour. Find the probability that fewer than 3 emails arrive in a 10-minute period.

\(\lambda = 9 \times \dfrac{10}{60} = 1.5\), so \(X \sim \text{Po}(1.5)\). ✓
\(\text{P}(X < 3) = \text{P}(X \le 2) = e^{-1.5}\left(1 + 1.5 + \dfrac{1.5^{2}}{2}\right) = 3.625e^{-1.5}\). ✓
\(= 0.809\) (3 s.f.). ✓

Analysis

[4 marks]

The numbers of typing errors on 100 pages were counted.

Errors	0	1	2	3	4	5
Pages	14	27	27	18	9	5

Calculate the mean and variance of the number of errors per page, and explain whether a Poisson model is suitable.

\(\sum fx = 0 + 27 + 54 + 54 + 36 + 25 = 196\), so the mean is \(\dfrac{196}{100} = 1.96\). ✓
\(\sum fx^{2} = 0 + 27 + 108 + 162 + 144 + 125 = 566\), so the variance is \(\dfrac{566}{100} - 1.96^{2} = 5.66 - 3.8416 = 1.82\). ✓
The mean and variance are close, which supports a Poisson model with \(\lambda \approx 1.96\), provided errors occur independently and at random. ✓

Analysis

[4 marks]

\(X \sim \text{Po}(\lambda)\) and \(\text{P}(X = 2) = \text{P}(X = 3)\).

(a) Show that \(\lambda = 3\). (b) Hence find \(\text{P}(X \ge 2)\).

(a) \(e^{-\lambda}\dfrac{\lambda^{2}}{2} = e^{-\lambda}\dfrac{\lambda^{3}}{6}\); dividing by \(e^{-\lambda}\lambda^{2}\) gives \(\dfrac{1}{2} = \dfrac{\lambda}{6}\), so \(\lambda = 3\). ✓
(b) \(\text{P}(X \ge 2) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} = 1 - 0.1991 = 0.801\). ✓

Analysis

[6 marks]

Requests to hire a car from a small firm occur at random at a mean rate of 2.4 per day. The firm has 4 cars.

(a) Find the probability that, on a given day, the demand for cars cannot be met.

(b) Find the probability that the demand cannot be met on at least one day of a 5-day week.

(c) Find the least number of cars the firm should have so that the probability that demand cannot be met on a given day is less than 0.05.

(a) \(X \sim \text{Po}(2.4)\). Demand is not met when \(X > 4\): \(\text{P}(X \ge 5) = 1 - \text{P}(X \le 4)\). ✓
\(\text{P}(X \le 4) = e^{-2.4}\left(1 + 2.4 + 2.88 + 2.304 + 1.3824\right) = 9.9664e^{-2.4} = 0.9041\), so \(\text{P}(X > 4) = 0.0959\). ✓
(b) \(1 - (1 - 0.0959)^{5} = 1 - 0.9041^{5} = 1 - 0.6042 = 0.396\). ✓
(c) With \(n\) cars we need \(\text{P}(X > n) < 0.05\). \(n = 4\): \(0.0959\) (too big). \(n = 5\): \(\text{P}(X > 5) = 0.0357 < 0.05\). ✓
The firm needs at least 5 cars. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.