Question

For each day, independent of the others, the length of time for one individual to be servedat 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

First, we find the probability that a person is served in less than 3 minutes on a single day, using the exponential distribution.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:  

$f(x)=\mu e^{-\mu x}$

in which $\mu=1/m$ is the decay parameter.

The probability that x is lower or equal to a is given by:  

$P(X\le x)=1-e^{-\mu x}$

In this problem, mean of 4 minutes, hence $m=4, \mu=0.25$

The probability that a person is served in less than 3 minutes on a single day is:

$P(X\le 3)=1-e^{-0.25(3)}=0.5276$

Now, for the 6 days, we use the binomial distribution:

$P(X=x)=C_{n,x}p^x(1-p)^{n-x}$

$C_{n,x}=\frac {n!} {x!(n-x)!}$

The parameters are:

• x is the number of successes.
• n is the number of trials
• p is the probability of a success on a single trial.

In this problem $n=6$ , $p=0.5276$ .

Hence, $P(X\ge4)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=C_{6,4}*(0.5276)^4*(0.4724)^2=0.2594$

$P(X=5)=C_{6,5}*(0.5276)^5*(0.4724)^1=0.1159$

$P(X=6)=C_{6,6}*(0.5276)^6*(0.4724)^0=0.0216$

Then, $P(X\ge 4)=0.2594+0.1159+0.0216=0.3969\approx39.7$ % that a person is served in less than 3 minutes on at least 4 of the next 6 days.