A charity issues a large number of certificates each costing $£10$ and each being repayable one year after issue. Of these certificates, $1$% are randomly selected to receive a prize of £10 such that they are repaid as $£20$. The remaining $99$% are repaid at their face value of $£10$.

Consider a person who purchases $200$ of these certificates.

Use a Poisson approximation to this binomial distribution to approximate the probability that this person is repaid more than $£2,040$.

My attempt,

I know that $N$~$Bin(200,0,01)$ can be approximated to $Poi(2)$

So I've to calculate $P(S>2040)$, but I don't know how to proceed.

Hope someone can explain it to me. Thanks in advance.

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ Commented Nov 6, 2017 at 8:19

1 Answer 1


First, the Poisson approximation states $Bin(n,p)\simeq Poi(np)$ so we can use $Poi(2)$.

Now, The basic repay is £10, each prize is additional £10 so we'd like to inquire the probability of $k>4$ prizes, this can be found using $P(k>4)=1-P(k\leq4)$.

$$P(k\leq4)=P(k=0)+P(k=1)+P(k=2)+P(k=3)+P(k=4)=e^{-2}(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!})=e^{-2}(\frac{1}{1}+\frac{2}{1}+\frac{4}{2}+\frac{8}{6}+\frac{16}{24})=0.135335\cdot7=0.947347\Rightarrow P(k>4)=1-P(k\leq4)=0.052653$$

So the probability of gaining more than £2040 is 0.052653. Two thing you should note:

  1. The Poisson approximation lets you work with the number of occurred events regardless of their order of appearance within the sequence of $n$ events.

  2. This approximation assumes $n\rightarrow\infty,p\rightarrow0$. While everything holds in this question (in the manner of 'we did not get anything odd', as is the usual case with textbook questions), I'd say that for $n=200$ this is a quite risky case of using this approximation.

  • $\begingroup$ I had thought they would get nothing except for the dollar 20 chance $\endgroup$
    – Deep North
    Commented Nov 6, 2017 at 9:25
  • $\begingroup$ @DeepNorth for this case I really can't see the benefit of using the approximation, except for possibly being easier to write with code. $\endgroup$
    – Spätzle
    Commented Nov 6, 2017 at 9:33
  • $\begingroup$ Not sure about the "riskiness" here. The exact answer matches to two decimal places in this case. It's the $p\approx 0$ part that mostly determines the accuracy of the approximation (one way to see this is the ratio of binomial to poisson variance is $1-p $ - nothing to do with $n $) $\endgroup$ Commented Nov 6, 2017 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.