Chris Tynan
Posted on Saturday, 08 February, 2003 - 09:04 pm:
STEP I, 2001, q13:

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

I decided that the number of plates broken had Po(4l) and the probability of the mathmo breaking n plates as having Po(l) as well as the number of plates broken by others having Po(3l). But I don't know where to proceed, since I've only just learned Poisson distribution.

Hints would be great...

Chris

Kerwin Hui
Posted on Saturday, 08 February, 2003 - 10:10 pm:
You can do this by just enumerating the possibilities. However, there is a slicker way by multinomial distribution (which is the obvious generalisation of binomial distribution). To save some messy calculations, note that a broken plate is equally likely to have been broken by A,B,C or D (the four mathematicians).

Kerwin
Chris Tynan
Posted on Sunday, 09 February, 2003 - 01:42 pm:
Kerwin, I don't really understand multinomial distribution, could you explain please?

Also, the possibilities appear to me to be too great for enumeration, or perhaps I'm missing something obvious...

Chris
Chris Purcell
Posted on Monday, 10 February, 2003 - 12:00 pm:
Five plates, four people. That's a lot of choices - except you only care about the mathematician, giving you a total of 6 ways the 5 plates could be broken. Can you see how to proceed from there using independence and conditional probabilities?

Chris
Chris Tynan
Posted on Monday, 10 February, 2003 - 07:45 pm:
Calling the number plates broken by other students O (O ~ Po(3l)) and the amount broken by the mathematician M (M ~ Po(l)), considering the cases when O=2 M=3, O=1, M=4 and O=0 M=5, I obtain P(M ³ 3)=e 4ll5 106/120.

Now, using your hint about conditional probability, P(M ³ 3 | X=5) (calling the number of plates broken X (X ~ Po(4l)) =P(M ³ 3 and X=5)/P(X=5)=e4ll5 106/120.5!/ e4l(4l)5 which gives the required probability I think.

Many Thanks

Chris