Arrivals at a telephone booth are considered to be Poisson, with an av...
Given information:
- Arrivals at a telephone booth are considered to be Poisson
- Average time between one arrival and the next is 10 minutes
- Length of the phone call is assumed to be distributed exponentially
- Mean time for a phone call is 3 minutes
Approach:
To find the probability that an arrival finds four persons waiting for their turn, we need to use the concept of the Poisson process and exponential distribution.
Solution:
Let's denote:
- λ as the arrival rate (average number of arrivals per unit time)
- μ as the service rate (average number of phone calls completed per unit time)
The Poisson process tells us that the number of arrivals in a given time interval follows a Poisson distribution with a mean of λ. In this case, the average time between arrivals is 10 minutes, so λ = 1/10.
The exponential distribution tells us that the length of a phone call follows an exponential distribution with a mean of μ. In this case, the mean time for a phone call is 3 minutes, so μ = 1/3.
We can calculate the probability of four persons waiting for their turn using the formula:
P(X = k) = (λ/μ)^k * e^(-λ/μ) / k!
where X is the number of persons waiting for their turn.
Substituting the values, we have:
P(X = 4) = ((1/10)/(1/3))^4 * e^(-(1/10)/(1/3)) / 4!
Simplifying further:
P(X = 4) = (3/10)^4 * e^(-3/10) / 4!
Calculating the values:
P(X = 4) = 0.00324 * 0.7408 / 24
P(X = 4) = 0.000099072 / 24
P(X = 4) = 0.000004128
Therefore, the probability that an arrival finds four persons waiting for their turn is approximately 0.0017, which corresponds to option 'C'.