The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. What I the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days?
To find the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days, we can use the exponential distribution and the concept of cumulative distribution function (CDF).
The exponential distribution is characterized by a single parameter λ, which is the inverse of the mean (μ). In this case, the mean is 4 minutes, so λ = 1/4.
The CDF of the exponential distribution is given by the formula:
CDF(x) = 1 - e^(-λx)
Where x is the value of the random variable.
To solve this problem, we need to calculate the probability of being served in less than 3 minutes on each of the 6 days, and then find the probability of having at least 4 successes.
Let's calculate the probability for one day first:
P(X < 3 minutes) = 1 - e^(-λx) = 1 - e^(-1/4 * 3) = 1 - e^(-3/4) ≈ 0.7135
Now, to find the probability of having at least 4 successes in 6 days, we can use the binomial distribution. The formula for the probability of having k successes in n trials is:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
In this case, n = 6 (6 days), p = 0.7135 (probability of being served in less than 3 minutes on a single day), and k ranges from 4 to 6.
P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6)
= (6 choose 4) * (0.7135)^4 * (1 - 0.7135)^(6 - 4)
+ (6 choose 5) * (0.7135)^5 * (1 - 0.7135)^(6 - 5)
+ (6 choose 6) * (0.7135)^6 * (1 - 0.7135)^(6 - 6)
By plugging in the values and doing the calculations, we get:
P(X >= 4) ≈ 0.0363 + 0.0064 + 0.0004 ≈ 0.0431
So, the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days is approximately 0.0431.