Step 1 From provided information, An average of 15 particles per minute occurs, Mean of poisson distribution (m)=15 Process following poisson distribution and probability mass function is, P (X=x)= (e^(-m)m^(x))/(x!) a) The required probability of exactly 20 particles arrive in one minute of period can be obtained as: P (X=20)= (e^(-15)15^(20))/(20!)=0.04181 Therefore, probability that exactly 20 particles arrive in one minute of period is 0.04181. b) The required probability of exactly one particle arrives in one second period can be obtained as: One minute=15 particles 60 seconds=15 particles 1 second=0.25 particle So, here m=0.25 P (X=1)= (e^(-0.25)0.25^(1))/(1!)=0.19470 Therefore, probability that exactly one particle arrives in one second period is 0.19470. c) The required probability that at least one particle arrives in particular one second period can be obtained as: P (X>=1)=1-P (X=0) =1- (e^(-0.25)0.25^(0))/(0!) =0.80529 Therefore probability that at least one particle arrives in particular one second period is 0.80529. 1. d) The required probability that at least two particles arrive in particular four second period can be obtained as: in 1 second 0.25 particle arrives, so in 4 seconds 1 particle will arrive. Therefore m=1, P (X>=2)=1- [P (X=0)+P (X=1) ] =1- [ ((e^(-1)1^(0))/(0!))+ ((e^(-1)1^(1))/(1!)) ] =0.26426 So, probability that at least two particles arrive in particular four second period is 0.26426.