Suppose the university wifi has \"technical issues\" according to a Poisson proc

Question

Suppose the university wifi has "technical issues" according to a Poisson process with rate 4 per day (so per 24 hours) (a) 2 points What is the probability that there is at least one failure event in the next one hour (b) 2 points What is the probability that there is exactly one failure event between 8:00 am and 10:00 am tomorrow and exactly one failure event between 3:00 pm and 4:30 pm. Note: here you are using assumption 3 on slide 32 in Lecture 7, namely independence of things that happen in different non-overlapping intervals of time.

Explanation / Answer

Ans:

Poisson distribution:

There are on average 4 failure evet per day,so 4/24 =1/6 failure per hr.

a)

P(atleast one failure)=1-P(no failure)=1-P(X=0)

=1-e-1/6

=1-0.8465

=0.1535

b)Exactly one failiure between 8 to 10 am i.e 2 hrs interval,so one average 2/6=1/3 failure per 2 hr

P(x=1)=e-1/3*((1/3)1/1!)=0.2388

Exactly one failure betweeb 3 to 4.30 pm i.e 1.5 hr interval,so on average 1.5/6=0.25 failure per 1.5 hr.

P(x=1)=e-0.25*(0.251/1!)=0.1947

As,above two events are independent,so probability of happenning bothe vents together

=0.2388*0.1947=0.0465

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