To find the probability that it will take more than 6 minutes for all the customers currently in line to check out, we need to use the exponential distribution.
Let X be the time it takes for one customer to check out. We know that X follows an exponential distribution with a mean of 3 minutes (since each customer takes 3 minutes to check out). Therefore, the parameter lambda of the exponential distribution is lambda = 1/3.
Let Y be the total time it takes for all the customers currently in line to check out. If there are n customers in line, then Y follows a gamma distribution with parameters n and lambda (since the sum of n independent exponential random variables with parameter lambda follows a gamma distribution with parameters n and lambda).
We want to find the probability that Y is greater than 6 minutes. That is:
P(Y > 6) = 1 - P(Y ≤ 6)
Using the cumulative distribution function (CDF) of the gamma distribution, we can compute:
P(Y ≤ 6) = GammaCDF(n, lambda, 6)
where GammaCDF is the CDF of the gamma distribution with parameters n and lambda.
Therefore, we have:
P(Y > 6) = 1 - GammaCDF(n, lambda, 6)
For example, if there are currently 5 customers in line, then we have:
lambda = 1/3
n = 5
P(Y > 6) = 1 - GammaCDF(5, 1/3, 6) = 0.016
So, the probability that it will take more than 6 minutes for all the customers currently in line to check out is 0.016 or approximately 1.6%.