Busy people arrive at the park according to a Poisson process with rate λ1=3/hour. Relaxed people arrive at the park according to an independent Poisson process with rate λ2=2/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?
(2C2)*(2/(2+3))²=4/25
Well, let's take a stroll through the park of probability, shall we?
The Poisson process describes the arrival of events over time, where the events occur independently and at a constant rate. In this case, busy people arrive at a rate of λ1 = 3 per hour, and relaxed people arrive at a rate of λ2 = 2 per hour.
Now, we're interested in finding the probability that both people who arrived during the last 10 minutes are relaxed.
First, let's convert the rates to a per-minute basis. We have λ1 = 3/60 = 1/20 per minute for busy people, and λ2 = 2/60 = 1/30 per minute for relaxed people.
Given that exactly two people arrived during the last 10 minutes, we have a total of three possibilities:
1. Both busy people arrived.
2. Both relaxed people arrived.
3. One busy person and one relaxed person arrived.
Now, let's calculate each possibility:
1. The probability of both busy people arriving is (1/20)^2 = 1/400.
2. The probability of both relaxed people arriving is (1/30)^2 = 1/900.
3. The probability of one busy person and one relaxed person arriving is 2 * (1/20) * (1/30) = 1/300.
To find the probability that both people who arrived are relaxed, we sum up the probabilities of possibilities 2 and 3:
P(both people relaxed) = P(both relaxed) + P(one busy and one relaxed)
= 1/900 + 1/300
= 1/900 + 3/900
= 4/900
= 1/225.
So, the probability that both people who arrived during the last 10 minutes are relaxed is 1/225.
Now, that's a pretty relaxing probability, isn't it?
To find the probability that both people who arrived at the park during the last 10 minutes are relaxed, we need to use conditional probability.
Let's denote the event "two people arrived at the park in the last 10 minutes" as A and the event "both people are relaxed" as B.
We know that A occurred, i.e., exactly two people arrived in the last 10 minutes.
Now, we need to find the probability of event B given that event A occurred, denoted as P(B|A).
Using Bayes' theorem, we have:
P(B|A) = (P(A|B) * P(B)) / P(A)
P(A|B) is the probability that exactly two people arrived in the last 10 minutes given that both people are relaxed. Since relaxed people arrive according to a Poisson process with rate λ2=2/hour, the probability of exactly two relaxed people arriving in the last 10 minutes is given by the Poisson probability mass function:
P(A|B) = (λ2*t)^n * exp(-λ2*t) / n!
Here, λ2 = 2/hour, t = 10/60 hours (10 minutes converted to hours), and n = 2 (exactly two people arrived).
P(B) is the probability that both people are relaxed. Since relaxed people arrive according to a Poisson process with rate λ2=2/hour, the probability of one relaxed person arriving in the last 10 minutes is given by the Poisson probability mass function with λ2*t = 2/6 (the rate multiplied by the time period):
P(B) = (λ2*t)^1 * exp(-λ2*t) / 1!
Finally, P(A) is the probability that exactly two people arrived in the last 10 minutes, regardless of their relaxation status. This can be calculated by summing the probabilities of two people arriving, either both being relaxed or one being relaxed and the other being busy.
P(A) = P(A|B) * P(B) + P(A|~B) * P(~B)
Here, P(A|~B) is the probability that exactly two people arrived in the last 10 minutes given that one is relaxed and the other is busy.
Since busy people arrive according to a Poisson process with rate λ1=3/hour, the probability of one busy person arriving in the last 10 minutes is given by the Poisson probability mass function with λ1*t = 3/6 (the rate multiplied by the time period):
P(A|~B) = (λ1*t)^1 * exp(-λ1*t) / 1!
P(~B) is the probability that one person is relaxed and the other is busy:
P(~B) = 2 * P(B) * P(~B) = 2 * (λ2*t)^1 * exp(-λ2*t) / 1! * (λ1*t)^1 * exp(-λ1*t) / 1!
Now, we can substitute these values into the equation to find P(B|A).
P(B|A) = (P(A|B) * P(B)) / P(A)
P(A|B) = (λ2*t)^n * exp(-λ2*t) / n! = (2/6)^2 * exp(-2/6) / 2!
P(B) = (λ2*t)^1 * exp(-λ2*t) / 1! = (2/6) * exp(-2/6) / 1!
P(A) = P(A|B) * P(B) + P(A|~B) * P(~B) = [(2/6)^2 * exp(-2/6) / 2!) * (2/6) * exp(-2/6) / 1!] + [(2/6) * exp(-2/6) / 1! * (3/6) * exp(-3/6) / 1!]
Finally, we can substitute these values into the equation to find P(B|A).
To find the probability that both people who arrived in the last 10 minutes are relaxed, we need to use the concept of conditional probability.
Let's denote the event that exactly two people arrived in the last 10 minutes as event A, and the event that both of them are relaxed as event B.
P(A) is the probability that exactly two people arrived in the last 10 minutes. This can be calculated using the Poisson distribution.
P(A) = e^(-λ) * (λ^k) / k!
Where λ is the average number of arrivals in a given time period, and k is the number of arrivals we are interested in.
In this case, λ = λ1 + λ2 = 3 + 2 = 5 arrivals per hour. Since we are looking at a 10-minute time period, we need to adjust the rate accordingly.
λ = 5 * (10/60) = 5/6 arrivals in 10 minutes
Using the formula, P(A) = e^(-5/6) * ((5/6)^2) / 2!
Next, we need to calculate P(B|A), which is the probability that both people are relaxed given that exactly two people arrived.
P(B|A) = P(A∩B) / P(A)
We know that P(A∩B) is the probability that both people are relaxed and exactly two people arrive. Since these events are independent, we can multiply their probabilities.
P(A∩B) = P(B) * P(A)
P(B) is the probability that a single person arriving in the last 10 minutes is relaxed. This can be calculated using the rate of relaxed people, λ2.
P(B) = λ2 / λ = 2 / 5
Substituting the values into the equation, P(A∩B) = (2 / 5) * P(A)
Finally, we can calculate P(B|A) using P(A∩B) and P(A).
P(B|A) = P(A∩B) / P(A) = [(2 / 5) * P(A)] / P(A)
P(B|A) simplifies to 2 / 5
Therefore, the probability that both people who arrived in the last 10 minutes are relaxed is 2/5 or 0.4.