The number of train passengers who fail to pay for their tickets in a certain region has a poison distribution with a mean of 4 per train

i) all passengers pay their ticket?
ii) More than 1 passenger fail to pay for their tickets

If there are 4800 trains per month in the region,

iii) Find the expected number of trains with at least 1 non paying passenger.
iv) Calculate the average cost to the train company during a month if the average cost of a ticket is $12.54 2. Originally Posted by whleow The number of train passengers who fail to pay for their tickets in a certain region has a poison distribution with a mean of 4 per train i) all passengers pay their ticket? ii) More than 1 passenger fail to pay for their tickets If there are 4800 trains per month in the region, iii) Find the expected number of trains with at least 1 non paying passenger. iv) Calculate the average cost to the train company during a month if the average cost of a ticket is$12.54
i)= P(X=0) = $e^-4$ = 0.01832
ii)= 1-P(X=1)-P(X=0) = $e^-4$(1+4) = 0.09157
iii)+iv) Not sure, this could mean I've done the previous ones wrong...

3. Originally Posted by Nyoxis
i)= P(X=0) = $e^-4$ = 0.01832
ii)= 1-P(X=1)-P(X=0) = $e^-4$(1+4) = 0.09157
iii)+iv) Not sure, this could mean I've done the previous ones wrong...
iii) X is the random variable number of non-paying passengers on a train.

X ~ Poisson(mean = 4).

Let Y be the random variable number of trains with at least one non-paying passenger.

Then Y ~ Binomial(n = 4800, p = 0.98168)

where $p = \Pr(X > 0) = 1 - \Pr(X = 0) \approx 1 - 0.01832 = 0.98168$.

Then E(Y) = np ( ) = .....

iv) (Answer to iii)(\$12.54) = .....