Na poisson probability problem, the rate of errors is one every two hours. what is the probability of at most three defects in four hours

Mean for each hour = 1
Mean for four hours, m = 4

P(x,m)=m^x*e^(-m)/x!
where x is number of defects.

P(X<=3,4)
=P(X=0,4)+P(X=1,4)+P(X=2,4)+P(X=3,4)
=0.01832+0.07326+0.14653+0.19537
=0.43355

Probability of at most 3 defects in four hours is 0.43355