I have a problem in solving a probability problem which does not seem to be hard to me but I can not get the correct answer.
Let $X,\tilde{X}\sim \operatorname{Exp}(\lambda)$, $Y\sim \operatorname{Exp}(\beta)$ and $Z=T-X\mid T>X$ where, $T$ is a fixed constant. Moreover, $\tilde{X},Y, Z$ are independent. I am interested in finding probability of
1) $P(Z=\min(Z,\tilde{X},Y))=P(Z<\tilde{X},Y)$
2) $P(Y=\min(Z,\tilde{X},Y))=P(Y<\tilde{X},Z)$
3) $P(\tilde{X}=\min(Z,\tilde{X},Y))=P(\tilde{X}<Y,Z)$
Please note that sum of these probabilities must be 1 as they are transition probabilities of a certain state in a stochastic process problem.
I know that distribution of $Z$ is truncated exponential. Hence, I wrote its pdf as $f(z)=\frac{\lambda e^{-\lambda(T-Z)}}{1-e^{-\lambda T}}$, $Z<T$. Therefore, for (1), I solved the integral
$$P(Z<\tilde{X},Y)=\int_o^T f(z)P(\tilde{X}>Z,Y>Z) \, dz= \int_o^T f(z) P(\tilde{X}>Z)P(Y>Z) =\int_o^T f(z)e^{-\lambda z}e^{-\beta z} \, dz = \frac{(1-e^{-T\beta})(\lambda e^{-\lambda T})}{\beta (1-e^{-\lambda T})}$$
Similarly I solved 2 and 3. After I got the probabilities for 1, 2 and 3, their sum did not equal to 1! I could not figure out where I am wrong. I would be happy if anyone help me in solving this problem.