GATE 2024 IN Q.65: solution and answer

Official answer: 0.130 to 0.140Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Check the density: ${\int }_0^{\infty }{e}^{-x}dx={\left[-e^{-x}\right]}_0^{\infty }=0-(-1)=1$, and $f_X(x)\ge 0$ everywhere, so $f_X$ is a valid pdf. It is the exponential distribution with rate $\lambda =1$.

2. The probability is the area under the pdf to the right of 2: $$P(X>2)={\int }_2^{\infty }{f}_X(x)dx={\int }_2^{\infty }{e}^{-x}dx$$

3. Evaluate: $${\int }_2^{\infty }{e}^{-x}dx={\left[-e^{-x}\right]}_2^{\infty }=0-\left(-e^{-2}\right)=e^{-2}$$

4. Cross-check with the exponential survival function $P(X>a)=e^{-\lambda a}$: with $\lambda =1$ and $a=2$ it gives $e^{-2}$ again.

5. Numerically, $e^{-2}=0.135335\dots$, so $P(X>2)\approx 0.135$ (rounded off to three decimal places).