Assume emails arrive in your inbox modeled by a poisson proc

Assume emails arrive in your inbox modeled by a poisson process with a mean of 4 per hour. Let X denote the the time unitl your next e-mail, which is therefore exponentially distributed.
a) mean time until the next email you recieve
b) mean number of email you expect to get in the next 5 days
c)probability that you DO NOT get an email in the next 5 minutes
d)mean number of emails you would expect in the next 20 minutes
e)probablity you get an email in the next 20 min

Solution

a) mean time until the next email you recieve

1/4=0.25

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b) mean number of email you expect to get in the next 5 days

Given X follows Poisson distribution with mean=4*24*5=480

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c)probability that you DO NOT get an email in the next 5 minutes

Given X follows Poisson distribbution with mean=4*5/60=0.333333 in 5 minutes

P(X=x)=(0.333333^x)*exp(-0.333333))/x!

So the probability is

P(X=0) =(0.333333^0)*exp(-0.333333))/1=0.7165315

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d)mean number of emails you would expect in the next 20 minutes

mean=4*20/60= 1.333333

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e)probablity you get an email in the next 20 min

P(X=x)=( 1.333333^x)*exp(- 1.333333)/x!

So P(X=1)=( 1.333333^1)*exp(- 1.333333)/1=0.3514629