Question

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 is the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days?

Answer 1

0.3968

Explanation:

Lets first calculate the probability of the individual to be server in less than 3 minutes. We will call X the amount of time the person had to wait.

`P_X(3) =  1/4* \int\limits^3_0 {e^(-0.25t)} \, dt = 1/4*(-4)*(e^(-0.25t) \, |_0^3) = -1(e^(-0.75)-1) = 1-e^(-0.75) = 0.5276`

We make this experiment, that has 0.5276 probability of success, 6 times. The total amount of success is a binomial distribution B(6,0.5276). Lets call this random variable Y. Note that

`P(Y \geq 4) = P(Y=4) + P(Y=5)+P(Y=6)`

`P_Y(4) = {6 \choose 4} * 0.5276^4 * (1-0.5276)^2 = 0.2594`

`P_Y(5) = {6 \choose 5} *0.5276^5*(1-0.5276) = 0.1158`

`P_Y(6) = 0.5276^6 = 0.0216`

Therefore,

`P(Y \geq 4) = P(Y=4) + P(Y=5)+P(Y=6) = 0.2594+ 0.1158+0.0216 = 0.3968`