Originally Posted by

**pantera** Breakdowns of the lifts in an office building at any given time are independent of one another and can be modelled using a Poisson Distribution with a mean 0.2 per day.

1. Determine the probability that there will be exactly 4 breakdowns during the month of June.

2. The probability that there are more than 3 breakdowns during the month of June.

3. The probability that there are no breakdowns during the first 5 days of June.

Mr F says: I think that you need to first calculate the probability of no breakdown on a single day: $\displaystyle {\color{red}p = \frac{0.2^0 e^{-0.2}}{0!}}$. Then probability of no breakdown in first five days is p^5.

4. The probability that the first breakdown in June occures on June 3rd.

Mr F says: p^2 (1 - p). See also geometric distribution.

5. It costs 1850 Euros to service the lifts when they breakdown. Find the expected cost of service for June.

Mr F says: The expected number of breakdowns in June is 6. Therefore ......

6. Determine the probability that there will be no breakdown in exactly 4 out of the first 5 days in June.

Mr F says: Let Y be the random variable number of breakdowns in first five days of June. Then Y ~ Binomial(n = 5, p = whatever you calculated from #3 above). Calculate Pr(Y = 4) ......

Thanks!