Question

A student receiving texts can be modeled as a Poisson process at a rate of 3 per hour.

a) What is the probability that the student receives 2 or more texts in the next 30 minutes?

b) Given that the student received 3 texts in the last 10 minutes, what is the probability that the next text will arrive within 15 minutes?

c) Among a group of 10 students, what is the probability that at least 2 will receive at least one text within 15 minutes?

Answer 1

ANSWER::

a)

expected number of text in 30 minutes =3*30/60=1.5

P(receives 2 or more text) =P(X>=2)=1-P(X<=1) =1-(P(X=0)+P(X=1))

=1-(e-1.51.50/0!+e-1.51.51/1!)

=0.4422

b)expected number of text in 15 minutes =3*15/60=0.75

here since probability of an event is independent from time interval to time interval

therefore P(next text arrive in 15 minutes) =1-P(no text in next 15 minutes)

=1-e-0.75*0.750/0!

=0.5276

c)

here this is binomial with parameter n=10 and p=0.5276

(at least 2 ) =P(X>=2)=1-P(X<=1)

=1-((10C0)*(0.5276)^0*(1-0.5276)^10+ (10C1)*(0.5276)^1*(1-0.5276)^9)

=0.9933

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