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$i)$the service time for each clerk is exactly(nonrandom) ten minutes?
$ii)$the service times are $i$ with probability $\frac{1}{3}$,
$i=1,2,3$?$iii)$the service times are exponential with mean $\frac{1}{\mu}$?
Let $T_1$,$T_2$ and $T_3$ the service times for $A,B,C$
$i)$ $P(T_1>t_2+t_3|t_1=t_2=t_3=10)=0$
I did not have to do this but that's okay, since this is quite intuitive
$ii)$Let $\Omega$ the space of possible values of $t_1,t_2,t_3$, then
$$t_1>t_2+t_3\Leftrightarrow t_1=3,t_2=1,t_3=1$$
$$P(T_1>t_2+t_3)=P(T_1=3)P(T_2=1)P(T_3=1)=\frac{1}{3}\frac{1}{3}\frac{1}{3}=\frac{1}{27}$$
$iii)$ Here I'm confuse anyone can help?
EDIT: The answer of $iii)$ is $\frac{1}{4}$ but I don't know how to get it
Best Answer
What you want to calculate is $P(T_2+T_3<T_1)$, where $T_i$'s are independent exponentially distributed random variables.
First, let's develop some formulas for continuous independent random variables:
$$ P(X < Y) = \int_{x<y} f_{X,Y}(x,y)\ dxdy = \int_{-\infty}^\infty\int_{-\infty}^y f_X(x)f_Y(y)\ dxdy = \int_{-\infty}^\infty F_X(y)f_Y(y)\ dy $$
$$F_{X+Y}(z) = \int_{x+y<z} f_{X,Y}(x,y)\ dxdy = \int_{-\infty}^\infty \int_{-\infty}^{z-y} f_X(x)f_Y(y)\ dxdy = \int_{-\infty}^\infty F_X(z-y)f_Y(y)\ dy$$
Using these, we can calculate $$F_{T_2+T_3}(t) = (1 - e^{-\mu t}(\mu t + 1))1_{[0,\infty\rangle}(t)$$ and finally $$P(T_2+T_3<T_1) = \frac 1 4.$$
Let me know if you need any details.
EDIT: There is a simpler way to do this, closely related to memorylessness of exponential distribution. Let $X$ be exponentially distributed, $Y$ independent of $X$ and $f_Y(y) = 0,\ y < 0$. Then $$\begin{align*} P(X>Y) &= 1 - P(X<Y) = 1 - \int_{-\infty}^\infty F_X(y)f_Y(y)\ dy \\ &= 1 - \int_{0}^\infty(1 - e^{-\mu y})f_Y(y)\ dy = \int_{-\infty}^\infty e^{-\mu y}f_Y(y)\ dy = {E}(e^{-\mu Y})\end{align*}$$ Now, assuming $X$ is exponentially distributed, and $X,Y,Z$ pairwise independent, we have $$\begin{align*} P(X> Y + Z) = {E}(e^{-\mu(Y+Z)}) = {E}(e^{-\mu Y}e^{-\mu Z}) = {E}(e^{-\mu Y}){E}(e^{-\mu Z}) = P(X>Y)(X>Z) \end{align*}$$ All we need to notice that in your exercise symmetry implies $P(X<Y) = P(X>Y) = \frac 1 2$ (and same for $Z$).
Yet another way is to directly write down integral:
$$P(X>Y+Z) = \int_{x>y+z} f_{X,Y,Z}(x,y,z)\ dxdydz = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{y+z}^\infty f_X(x)f_Y(y)f_Z(z)\ dxdydz \\= \int_{-\infty}^\infty\int_{-\infty}^\infty (1-F_X(y+z))f_Y(y)f_Z(z)\ dydz$$ and calculate.