Because there are 60 seconds in a minute, we have 60λ = 330. It follows that λ = 5 1 /2 . Since the interarrival times have an Exp(λ) distribution, the expected time between messages is 1/λ = 0.18...

Because there are 60 seconds in a minute, we have 60λ = 330. It follows that λ = 5 1 /2 . Since the interarrival times have an Exp(λ) distribution, the expected time between messages is 1/λ = 0.18 second.

2.The intensity of this process is λ = 5 per m2. The area of the strip is 2 · (1/20) = 1/10 m2. Hence the probability that no defects occur in the strip