With a mean of λ , the probability mass distribution (pmf) is given by
[tex]P(k,\lambda)=\lambda^k*e^(-\lambda)/k![/tex]
for λ = 4, and k<2 (i.e. k=0 or 1)
P(k<2, λ )=P(k=1, λ ) + P(k=1, λ )
[tex]=4^0*e^(-4)/4!+4^1*e^(-4)/1![/tex]
[tex]=4^0*e^(-4)/4!+4^1*e^(-4)/1![/tex]
=0.01832+0.07326
=0.09158 (to the fifth place of decimal)
Note: Poisson processes have no memory, so 2 calls in first hour has the same probability as 2 calls in any other hour.