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?
To find the probability that both people are relaxed, we need to calculate the conditional probability that they are both relaxed given that exactly two people arrived in the last 10 minutes.
Let's define the following variables:
λ₁ = rate of busy people arriving per hour = 3 per hour
λ₂ = rate of relaxed people arriving per hour = 2 per hour
t = time period in hours
Now, let's solve the problem step-by-step:
Step 1: Convert the time period to hours.
Since we are given that the last 10 minutes, we need to divide it by 60 to convert it to hours.
10 minutes / 60 = 1/6 hour
Step 2: Calculate the rate of arrival during the last 10 minutes.
To calculate the rate of arrival during the last 10 minutes, we multiply the rate per hour by the time period in hours.
Rate during the last 10 minutes = λ * t
Rate during the last 10 minutes = 2 * (1/6) = 1/3
Step 3: Calculate the probability of exactly two people arriving during the last 10 minutes.
To calculate the probability of exactly two people arriving during this time period, we use the Poisson distribution formula.
Probability of a specific number of arrivals = (e^(-λ) * λ^k) / k!
Here, λ is the rate during the last 10 minutes, and k is the number of arrivals.
Probability of exactly two people arriving = (e^(-(1/3)) * ((1/3)^2)) / 2!
Probability of exactly two people arriving = (e^(-(1/3)) * (1/9)) / 2
Probability of exactly two people arriving = (e^(-(1/3)) * (1/9)) / 2 = (1/3) * (e^(-1/3)) = 0.170
Step 4: Calculate the probability that both arrivals are relaxed.
To find the probability that both arrivals are relaxed, we need to multiply the probability of exactly two arrivals by the ratio of relaxed arrivals to the total arrivals.
Probability that both arrivals are relaxed = (probability of exactly two arrivals) * (rate of relaxed arrivals / rate of total arrivals)
Probability that both arrivals are relaxed = 0.170 * (2 / (3 + 2))
Probability that both arrivals are relaxed = 0.170 * (2/5)
Probability that both arrivals are relaxed = 0.068
Therefore, the probability that both people are relaxed given that exactly two people arrived in the last 10 minutes is approximately 0.068, or 6.8%.
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.