Mathematics 9709/71 — October/November 2013

Walkthrough

We are told that each computer contains 1000 components and that on average 1 in 30 000 components is defective. If we let $X$ be the number of defective components in one computer, then $X$ could be modelled by a binomial distribution:

$$X \sim B\left(1000, \frac{1}{30000}\right)$$

However, calculating the binomial probability for at least one defective component would be awkward. Because $n = 1000$ is large and $p = \frac{1}{30000}$ is small, we can use the Poisson approximation to the binomial distribution. The Poisson parameter is the mean:

$$\lambda = np = 1000 \times \frac{1}{30000} = \frac{1}{30}$$

So $X \approx Po\left(\frac{1}{30}\right)$.

The event "at least 1 faulty component" is the complement of "no faulty components". Therefore:

$$P(X \ge 1) = 1 - P(X = 0)$$

For a Poisson distribution, $P(X = 0) = e^{-\lambda}$, so:

$$P(X \ge 1) = 1 - e^{-\frac{1}{30}}$$

Evaluating this gives $0.0328$ to 3 significant figures.

Key Takeaways

• The Poisson distribution is a suitable approximation to the binomial distribution when $n$ is large and $p$ is small.
• The mean of the approximating Poisson distribution is $\lambda = np$.
• For a Poisson distribution, $P(X = 0) = e^{-\lambda}$.
• "At least one" is often best handled using the complement rule: $P(X \ge 1) = 1 - P(X = 0)$.

Common Mistakes

• Using $\lambda = \frac{1}{30000}$ instead of $\lambda = np = \frac{1}{30}$. The given fraction is the probability per component, not the mean number of defects per computer.
• Calculating $P(X = 1)$ instead of $P(X \ge 1)$.
• Forgetting to use the complement rule and trying to sum infinitely many Poisson probabilities.
• Not showing the method. The mark scheme allows only B2 for a correct unsupported answer, so full working is needed for full marks.

Things to Be Careful About

• The final answer should be given to 3 significant figures: $0.0328$.
• The mark scheme awards B1 for the correct value of $\lambda$, M1 for using $1 - P(X = 0)$ with a Poisson distribution, and further marks for the correct evaluation. Show each of these steps clearly.
• If the binomial distribution is used instead, the special rule in the mark scheme awards only B2 for a correct final answer, because the question asks for a suitable approximate distribution, which is Poisson.
• Make sure the probability is between 0 and 1; $0.0328$ is a plausible small probability because the mean number of defects is only $\frac{1}{30}$.