Having trouble with the last part of this question. Not sure how the man would divide his pile of vouchers? It seems that you could interpret this question in a lot of ways. Any tips would be appreciated. Thanks :)

A petrol company issues a voucher with every 10 litres of petrol that a customer buys. Customers who send 50 vouchers to Head Office are entitled to a free gift but only one gift is allowed per household. After this promotion has been running for some time the company receives several hundred bundles of vouchers in each day’s post. It would take a long time, and hence not be cost effective, for someone to count each bundle and so they decide to weigh them instead. Working on the basis that the weight of a single voucher is a Normal variable with mean 36mg and standard deviation 2.5mg they decide to calculate the weight which will be exceeded by 95% of the bundles and only count the numbers of vouchers in bundles which weigh less than this calculated amount. A man has 150 vouchers, divides these into 3 bundles without checking the number in each bundle carefully, and sends one claim for a gift in his own name and two claims in the names of friends. Assess the probability that he will receive all three free gifts using relevant calculations. Give details of any additional assumptions that you have made.


Here are some hints to get you going...note that the last part of the question is forcing you to come up with some assumptions, so there isn't likely one answer:

The weight of the "reference pile" ($W_r$) used by the company will be distributed:

$$P_r\sim \mathcal{N}(50\times 36,50\times 2.5^2)=\mathcal{N}(1800,312.5)\implies \sigma_{W_r}\approx17.7$$

Therefore, their lower 5% cutoff is simply $1800-1.64\sigma\approx1,771$

Now, the man is making three piles which will contain$N_1,N_2,N_3$ cards such that $N_1+N_2+N_3=150$.

The weights of the piles will be $W_1,W_2,W_3$, where $W_i\sim \mathcal{N}(36N_i,2.5^2N_i)$

Thus, you need to come up with an assumption on the random process that generates the piles, then the probability of getting all three prizes is:

$$P(\cap_{i=1}^3W_i\geq1,771)=\sum P(N_1=n_1,N_2=n_2,N_3=n_3)P(\cap_{i=1}^3W_i\geq1,771|N_1=n_1,N_2=n_2,N_3=n_3)$$

The key to this problem is coming up with a model for $(N_1,N_2,N_3)$ that is both reasonable and simple....otherwise you'll have a real task ahead of you.

Welcome to the world of probabilistic modelling.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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