Question

Let Z be an exponential random variable with mean 1. That is, the cumulative distribution function of Z is given by
$F_z\left(x\right)=(\begin{array}{cc}1-e^{-x}& \mathrm{if}x\ge 0\\ 0& \mathrm{if}x<0\end{array}$
Then Pr(Z > 2 | Z > 1), rounded off to two decimal places, is equal to .

• A. 0.367

Answer 1

The correct option is A 0.367

Cumulative distribution function of z is

$F_Z\left(x\right)=(\begin{array}{cc}1-e^{-x}& \mathrm{if}x\ge 0\\ 0& \mathrm{if}x<0\end{array}$
$\mathrm{Pdf}{f}_Z\left(x\right)=\frac{{\mathrm{dF}}_Z\left(x\right)}{\mathrm{dx}}$

$F_z\left(x\right)=(\begin{array}{cc}{e}^{-x}& \mathrm{if}x\ge 0\\ 0& \mathrm{if}x<0\end{array}$

$\rho \left(\frac{z>2}{z>1}\right)=\frac{\rho \left(\left(z>2\right)\cap \left(z>1\right)\right)}{\rho (z>1)}$

$=\frac{\rho (z>2)}{\rho (z>1)}$

$\rho (z>2)={\int }_2^{\infty }{e}^{-x}\mathrm{dx}=e^{-2}$

$\rho (z>1)={\int }_1^{\infty }{e}^{-x}\mathrm{dx}=e^{-1}$

$\mathrm{hence}\rho \left(\frac{z>2}{z>1}\right)=\frac{\rho (z>2)}{\rho (z>1)}=\frac{e^{-2}}{e^{-1}}=e^{-1}=0.367$