As in an earlier exercise, busy people arrive at the park according to a Poisson process with rate lambda_1=3 per hour. Relaxed people arrive at the park according to an independent poisson process with rate lambda_2=2 per hour. Assume that no other people arrive at the park. During the last 10 minutes, exactly two people arrived at the park. What is the probability that they are both relaxed?

1 answer

To solve this problem, we can use Bayes' theorem. Let A be the event that both people are relaxed, and B be the event that exactly two people arrived in the last 10 minutes. We want to find P(A|B), the probability that both people are relaxed given that exactly two people arrived in the last 10 minutes.

We will need to use the following information:

1. P(A) = the probability that a given person is relaxed = lambda_2 / (lambda_1 + lambda_2) = 2 / (3 + 2) = 2/5
2. P(A') = the probability that a given person is busy = 1 - P(A) = 1 - 2/5 = 3/5
3. P(B|A) = the probability of exactly two people arriving in the last 10 minutes, given that they are both relaxed. This will follow a Poisson distribution with rate lambda_B_A = 2 * (10 minutes / 60 minutes per hour) * lambda_2 = 2 * (1/6) * 2 = 2/3. Therefore, P(B|A) = (e^(-lambda_B_A) * (lambda_B_A)^2) / 2! = (e^(-2/3) * (2/3)^2) / 2 = (4e^(-2/3)) / 9 ≈ 0.261

To calculate P(B), the probability of exactly two people arriving in the last 10 minutes, we can see that it is the sum of the probabilities of exactly two busy people or exactly two relaxed people:

P(B) = P(B|A')P(A') + P(B|A)P(A) = (e^(-lambda_B_A') * (lambda_B_A')^2) / 2! * (3/5) + (4e^(-2/3)) / 9 * (2/5)

Where lambda_B_A' = 2 * (10 minutes / 60 minutes per hour) * lambda_1 = 2 * (1/6) * 3 = 1

Plugging in these values, we get:

P(B) = (e^(-1) * 1^2) / 2! * (3/5) + (4e^(-2/3)) / 9 * (2/5) ≈ 0.224 + 0.174 ≈ 0.398

Finally, we can use Bayes' theorem to find P(A|B):

P(A|B) = (P(B|A) * P(A)) / P(B) = ((4e^(-2/3)) / 9 * (2/5)) / 0.398 ≈ 0.261 / 0.398 ≈ 0.654

Therefore, the probability that both people who arrived in the last 10 minutes are relaxed is approximately 0.654.