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Problem 68:  Breaking plates () 2001 Paper I

Four students, one of whom is a mathematician, take turns at washing up over a long period of time. The number of plates broken by any student in this time obeys a Poisson distribution, the probability of any given student breaking n plates being eλλnn! for some fixed constant λ, independent of the number of breakages by other students. Given that five plates are broken, find the probability that three or more were broken by the mathematician.


The way this is set up, it is largely a counting exercise (but see postmortem). To start with, you work out the probability of five breakages, then follow that with the probability that the mathematician broke three of more plates. You need to calculate the number of ways that 5 plates can be shared amongst four students, for which you have to consider each partition of the number 5 and the number of different ways it can arise.

Solution to problem 68

First we work out the probability of five breakages.

Let P(5,0,0,0) denote the probability that student A breaks 5 plates and students B, C and D break no plates. Then

P(5,0,0,0) = (Prob. of breaking 5) × (Prob of breaking none)3 = eλλ5 5! eλ 3 = λ5e4λ 5! .

The probability that one student breaks all the plates is 4 × λ5e4λ5!, the factor 4 because it could be any one of the four students.

Let P(4,1,0,0) denote the probability that student A breaks 4 plates, student B breaks one plate and students C and D break no plate. Then

P(4,1,0,0) = (Prob. of breaking 4) × (Prob. of breaking one) × (Prob of breaking none)2 = eλλ4 4! eλλ 1! eλ 2 = λ5e4λ 4! .

The probability that any one student breaks four plates and any other student breaks one plate is therefore 4 × 3 × λ5e4λ4!.

Considering P(3,2,0,0), P(3,1,1,0), P(2,2,1,0) and P(2,1,1,1) in turn, together with the number of different ways these probabilities can occur, shows that the probability of five breakages is

λ5e4λ 4 5! + 12 4!1! + 12 3!2! + 12 3!1!1! + 12 2!2!1! + 4 2!1!1!1! = λ5e4λ 1024 5! .

The probability that the mathematician (student A, say) breaks three or more is P(5,0,0,0) + 3P(4,1,0,0) + 3P(3,2,0,0) + 3P(3,1,1,0), i.e.

λ5e4λ 4 5! + 3 4!1! + 3 3!2! + 4 3!1!1! = λ5e4λ 106 5! .

The probability that the mathematician breaks three or more, given that five are broken is therefore 1061024.


If you did the question by the method suggested above, you will probably be wondering about the answer: why is it independent of λ; and why is the denominator 45. You will quickly decide that the Poisson distribution was a red herring (though I promise you that it was not an intentional herring), since there is no trace of the distribution in the answer.

The denominator is pretty suggestive. A completely different approach is as follows, It is clear that the probability that the mathematician breaks any given plate is 1 4. The probability that he or she breaks k plates is therefore binomial:

5 k 1 4k 3 45k

and adding the cases k = 3, k = 4 and k = 5 gives rather rapidly

90 + 15 + 1 1024 .

For the method given in the solution, we could substitute Pk for eλλkk! (as the probability of breaking k plates) and the calculation would work just the same, with all the Pks disappearing; try it.