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]
