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Suppose that $X$ and $Y$ are exponential r.v.'s with parameters $\lambda$ and $\mu$. Given the condition that $X$ is bounded by $Y$ (i.e., $X<Y$), how do I find the probability that a number $t$ falls between $X$ and $Y$?
Can I interpret this question as $P(X<t<Y\mid X<Y)$? If correct, how to compute it?
A follow-up question. Given the condition that $X<Y$ and $t<Y$, how do I compute the probability that $X<t$?
Your interpretation seems fine to me. See that: $$ P(X<Y\mid X<t<Y)=1. $$ Hence: $$ P(X<Y,X<t<Y)=P(X<t<Y)=P(X<t)P(Y>t)=(1-e^{-\lambda t})e^{-\mu t}. $$
Regarding your second question. Again remember that: $$ P(Y>X\mid Y>t)=1-\frac{\mu}{\lambda+\mu}e^{-\lambda x}. $$ So: $$ P(X<t\mid t<Y,X<Y)=\frac{P(X<Y,X<t<Y)}{P(X<Y,Y>t)}=\frac{e^{-\mu t}(1-e^{-\lambda t})}{e^{-\mu t}(1-\frac{\mu}{\lambda+\mu}e^{-\lambda t})}\\ =\frac{ (1-e^{-\lambda t})}{ (1-\frac{\mu}{\lambda+\mu}e^{-\lambda t})}. $$
