N a 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