The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are exactly 5 calls in one hour? (b) What is the probability that there are 3 or fewer calls in one hour? (c) What is the probability that there are exactly 15 calls in two hours? (d) What is the probability that there are exactly 5 calls in 30 minutes?

Answer with Step-by-step explanation:We know from Poisson distribution The probability that a random process with an average arrival rate of λ occurs 'n' times in time interval of 't' is given by[tex]P(n,t)=\frac{(\lambda t)^ne^{-\lambda t}}{n!}[/tex]Part a) The probability for 5 calls in 1 hour is [tex]P(5,1)=\frac{(10\times 1)^{5}e^{-10\times 1}}{5!}\\\\P(5,1)=0.0378[/tex]Part b) Probability of 3 or fewer calls occurs in 1 hour is the sum of the following probabilities 1) Only 1 call occurs in 1 hour.2) Only 2 calls occurs in 1 hour.3) Only 3 calls occurs in 1 hour.4) There is no cal in 1 hour.Thus we can write[tex]P(n<3,1)=\frac{(10\times 1)^0e^{-10}}{0!}+\frac{(10\times 1)^1e^{-10}}{1!}+\frac{(10\times 1)^2e^{-10}}{2!}+\frac{(10\times 1)^3e^{-10}}{3!}\\\\P(n<3,1)=0.0103[/tex]Part c)The probability for 15 calls in 2 hours is [tex]P(15,2)=\frac{(10\times 2)^{15}e^{-10\times 2}}{15!}\\\\P(15,2)=0.051[/tex]Part d)The probability for 5 calls in 30 minutes or  0.5 hours is [tex]P(5,0.5)=\frac{(10\times 0.5)^{5}e^{-10\times 0.5}}{5!}\\\\P(5,0.5)=0.175[/tex]