Probability question, poisson distribution?

Gil_Gunderson

Not sure of this probability question:

A waiter in a restaurant under-reports the number of tips he gets from customers. For every n customers who have tipped n-m tips are reported to the manager. The manager dines there once a day for n days. What are the chances that 3 of the tips given by the manager go unreported, if n=100 and m=10?

Not sure what technique to apply, I was thinking poisson distribution where lambda = 10 and k = 3. But this seems kind of simple, and the actual solution might be more complicated. Any help would be appreciated.

hivizman

In this situation, I think that the binomial distribution would be more appropriate. If you have n customers and m tips go unreported, then the maximum value that m can take is n (you can't have more tips being unreported than customers available to offer tips). The Poisson distribution assigns non-negative (though very small) probabilities to the situations where m is greater than n, so it doesn't fit the situation.

If you use the binomial distribution, then the probability that a tip is reported is (n-m)/n, or in the case given (1-0.1)/1 = 0.9. Assuming that the fact that it's the manager dining has no impact on this probability (which is a big and rather implausible assumption - I'd predict that, so long as the waiters recognise the manager, they would be more inclined to report the manager's tips), then the probability that three of the tips given by the manager out of 100 meals go unreported is:

(100!/(3!*97!))*(0.9^97)*(0.1^3)

This is a bit more cumbersome to calculate than the equivalent Poisson probability:

EXP(-10)*(10^3)/3!

By the way, I'm interpreting the question literally, that we are trying to find the probability that exactly three tips go unreported, rather than reading "three" as implying "three or fewer". But if you think that it's a trick question, then the answer is most likely zero - do you really think that the waiters won't report to the manager all of the manager's tips?

versc

Hi, Stuck on a poisson question myself.


The number of accidents occurring in a month is modelled as a homogeneous poisson process with rate lambda = 5 accidents per month.
Given that an accident has just occurred what is the probability that it will be more than 1 month until 3 further accidents have occurred?

Anyone any ideas?

Yakuza

The bit about "given an accident has just occured" is irrelevant. The Poisson distribution is memoryless, so the question is basically asking you what is the chance of there being exactly 3 accidents in the next month, given that there are 5 accidents a month. Remember that the probability of k events in a Poisson distribution with parameter [latex]\lambda[/latex] is [latex]\frac{{\lambda}^k \times e^{-\lambda}}{k!}[/latex]