Life of a virus has a probability distribution of $\lambda e^{-\lambda t}$. If it already lived for $n$ hours, how long it can be expected to live?
I kinda of know that without conditioning, $E(t)$ will be $1/\lambda$ by integrating. But with the condition added, I guess it should still be $1/\lambda$, however I don't know how to write formal proof for this.
Let $X$ denote the life of the virus in hours, with $X\sim \exp(\lambda)$. Let $Y:=X\mid X>n$ denote the life of the in hours given that the virus has lived $n$ hours. Then $$F_Y(y)=P(Y\le y)=\begin{cases}0, & \text{if } y\le n\\P(X\le y\mid X>n), &\text{if }y>n\end{cases}$$ where $$P(X\le y\mid X>n)=\frac{P(n<X\le y)}{P(X>n)}=\frac{F_X(y)-F_X(n)}{1-F_X(n)}$$ Hence, the probability density function of $Y$ is given by $$f_Y(y)=\frac{dF_Y(y)}{dy}=\frac{f_X(y)}{1-F_X(n)}=\frac{\lambda e^{-\lambda y}}{e^{-\lambda n}}=\lambda e^{-\lambda (y-n)}$$ for $y>n$. Now, you just need to compute the expected value of $Y$, using the usual formula $$\mathbb E[Y]=\int_{n}^{\infty}f_y(y)dy$$ which by the memoryless property of the exponential distribution will give you $1/\lambda$ as you correctly derived (from your intuition).