If $Y$ is a continuous random variable and $m$ is the median of the distribution, then $m$ is such that
$P(Y \leq m)=P(Y \geq m)=1 / 2 .$ If $Y_{1}, Y_{2}, \ldots, Y_{n}$ are independent, exponentially distributed random variables with mean $\beta$ and median $m$, Example 6.17 implies that $Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)$ does not have an exponential distribution. Use the general form of $F_{Y_{(v)}}(y)$ to show that $P\left(Y_{(n)}>m\right)=1-(.5)^{n}$