Question on probability and statistics?

Suppose a customer opening an account at a bank will either open a savings or checking account but not both. Assume that customers make decisions independently. From past experiences, customers who will open an account arrive according to a Poisson process. The number of customers (per day) arriving at the bank for opening a savings account is a Poisson random variable with parameter μ=7 and that for a checking account is also a Poisson random variable with parameter μ=3. Let X1 be the number of customers opening a savings account tomorrow, X2 be the number of customers opening a checking account tomorrow.


(a) What is the expected time between two successive customers opening an account?


(b) Given that n customers will come and open an account tomorrow, what is the conditional distribution of X1?


(c) Find the probability that the account a customer opens at the bank is a savings account.

Question on probability and statistics?
X1 ~ Poisson (7)


X2 ~ Poisson (3)





a) Let X be the number of accounts opened. X = X1 + X2 so


X ~ Poisson(10)





the waiting time between events that are distributed via the Poisson distribution follow the Exponential distribution with the same parameter or 1/parameter depending on how you want to define the distribution function. Let W be the wait time between account openings, W ~ Exp(λ = 10),





the density function is λ * e^(-λ *x) for x%26gt;0





the mean is 1/ λ so you expect every 10th of a day for an account to open.





(b) Find f(x1 | n) = f(x1 and n)/ f(n)





f(X1 = x | X1 + X2 = n)


= f(X1 = x and X2 = n -x) / f(X1 + X2 = n)





after you do the algebra you will find that


f(X1 | n) = n!/(x! (n-x)!) * ( λ1 / (λ1 + λ2))^x * ( λ2 / (λ1 + λ2))^(n - x)





where λ1 is the parameter for X1 and λ2 is the parameter for X2. If you don't recognize this, it is the binomial distribution.





f(X1 = x | n) = n!/(x! (n-x)!) * ( 0.7 )^x * ( 0.3 )^(n - x)











(c) the probability that a customer who comes into the bank opens a savings account is 0.7. look at (b) to see this.
Reply:If X1~Poi(μ1) and X2~Poi(μ2) are independent, then Y = X1 + X2 ~ Poi(μ1+μ2).


Therefore, the expected time between two customers is 1/(μ1+μ2) = 1/10th of a day (however long the bank's "day" is).





The distribution of X1 conditional upon Y=n is a binomial distribution:


X1|(Y=n) ~ Binom(n,μ1/(μ1+μ2)).





I'm not sure about the third part right now.