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.
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
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 that arrived in the past hour can be modeled by a Poisson distribution with parameter lambda_1*t, where t is the time in hours.
Since we are given that the observer visits the park at a specific time, let's say t=1 hour, we can calculate this probability using the Poisson PMF formula:
P(B=0) = (lambda_1*t)^0 * e^(-lambda_1*t) / 0!
= (3*1)^0 * e^(-3*1) / 0!
= e^(-3)
Therefore, the probability that B=0 is e^(-3).
To find the probability that B+R=1, we need to consider both the number of busy people (B) and the number of relaxed people (R) in the park at that moment. Since the arrivals of busy and relaxed people are independent processes, we can multiply their respective probabilities.
The probability of B=1 is given by the Poisson PMF formula:
P(B=1) = (lambda_1*t)^1 * e^(-lambda_1*t) / 1!
= (3*1)^1 * e^(-3*1) / 1!
= 3e^(-3)
The probability of R=0 is given by the Poisson PMF formula:
P(R=0) = (lambda_2*t)^0 * e^(-lambda_2*t) / 0!
= (2*1/2)^0 * e^(-2*1/2) / 0!
= e^(-1)
Therefore, the probability that B+R=1 is (3e^(-3)) * (e^(-1)) = 3e^(-4).
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