Busy people arrive at the park according to a Poisson process with rate lambda_1=3 per hour and stay in the park for exactly 1/6th of an hour. Relaxed people arrive at the park according to a poisson process with rate lambda_2=2 per hour and stay in the park for exactly half an hour. The arrivals of the busy and relaxed people are independent processes. An observer visits the park at a specific time and sees B busy and R relaxed people at the park at that moment.

Question

1. Find that probability that B=0. Hint: think about what must have happened in the immediate past. Recall also the formula for the poisson PMF with parameter lambda = (lambda^k*e^-lambda)/k factorial, for k=0,1,2,.....

2. Find the probability that B+R=1

Answer 1

1. To find the probability that B=0, we need to consider what must have happened in the immediate past. Since busy people arrive according to a Poisson process with rate lambda_1=3 per hour, the number of busy people at any given time also follows a Poisson distribution with the same rate.

Since the observer sees B busy people at the park, this means that there must have been B busy people who arrived before the observer's arrival. The number of busy people who arrived before the observer follows a Poisson distribution with rate lambda_1=3 per hour and a time span of 1/6th of an hour.

Therefore, the probability that B=0 is given by the Poisson PMF:

P(B=0) = (lambda_1*(1/6))^0 * e^(-lambda_1*(1/6)) / 0! = 1 * e^(-1/2) ≈ 0.6065

2. To find the probability that B+R=1, we need to consider all the different scenarios that can lead to B+R=1.

One scenario is that there are 0 busy people and 1 relaxed person at the park. From the previous question, we know that the probability of having 0 busy people is approximately 0.6065. Since relaxed people arrive according to a Poisson process with rate lambda_2=2 per hour, the number of relaxed people at any given time also follows a Poisson distribution with the same rate. Therefore, the probability of having 1 relaxed person is given by the Poisson PMF with lambda_2=2 and a time span of 1/2 an hour:

P(R=1) = (lambda_2*(1/2))^1 * e^(-lambda_2*(1/2)) / 1! = 2 * e^(-1) ≈ 0.7358

Therefore, the probability that B+R=1 is the product of the probabilities of having 0 busy people and 1 relaxed person:

P(B+R=1) = P(B=0) * P(R=1) ≈ 0.6065 * 0.7358 ≈ 0.4463