Question

Warnings of a possible virus attack on a particular laptop can be modeled as a Poisson process with 1 warning per hour. a) A person received 3 warnings in the last 1 hour of which 1 warning was 15 minutes back. What is the probability that this person has to wait for more than 3 hours for his next warning? b) In a 3-hour software update going on a laptop of 200 people, what is the probability that less than 4 of them will have no warnings during the course of the software update?

Answer 1

Answer:

a)

a.) Probability such that the person has to wait for more than 3 hrs for next warning

  Let P(E₁) be the probability of getting one warning per hour

   Let P(E₂) be the probability of getting two warning per hour

   Let P(E₃) be the probability of getting three warning per hour

   Let P(E₄) be the probability of getting four warning per hour

   and so on...

   Let P(E_n) be the probability of getting 'n' warning per hour

Therefore,

   Probability such that the person has to wait for more than 3 hrs for next warning = P(E₃)*(1-P(E₁))*(1-P(E₁))*(1-P(E₁))

   =P(E₃)*(1-P(E₁))³

b)

The required probability is =

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