I'm working on this problem on exponential distribution:
- Consider a post office with two clerks. Three people, A, B, and C, enter simultaneously. A and B go directly to the clerks, and C waits until either A or B leaves before he begins service. What is the probability that A is still in the post office after the other two have left when the service times are exponential with mean 1/μ?
I want to use this property on probability of inequality of exponential random variables
I want to sub in $T_A$, time that person A spends in the post office, for $X_2$ in the formula, and $T_B + T_C$ for $X_1$. And then I would simplify $T_B + T_C$ as $2T_B$, which by the property of exponential random variable would have a rate of μ/2. So I would get μ/2 in the numerator and μ/2 + μ in the denominator, which simplifies to 1/3. However, the solution I've found uses a different approach and gives a different result, for example #4 on here: https://www2.isye.gatech.edu/~sman/courses/6761/hw6f12solns.pdf. Can someone let me know why my method is incorrect?
The first step of your calculation is incorrect, the formula you have given only holds if $X_1$ and $X_2$ are exponentially distributed. However, $X_2:=T_B+T_C$ is not exponentially, but $\Gamma(2,1/\mu)-$distributed (you cannot just simplify $T_B+T_C$ as $2T_B$, the distribution of the sum of two independent random variables is given by its convolution). If you use the PDF of this distribution instead of the PDF of the exponential distribution in equation 1 of your source, the calculation should work.