For example: If P1 and P2 are exponential pdf's, with λ1 = 1, and λ2 = 2
$$P(X=x) = P^1(x) + P^2(x) - P^1(x)P^2(x)$$
We know E[Y] where Y follows an exponential distribution is $$1/λ$$
Would the expected value be: $$E[X] = 1/1 + 1/2 - (1/1)(1/2) = 1$$
No, the expectation operator is linear. It does not distribute
In this case, if $P_X(x)=P_1(x)+P_2(x)-P_1(x)P_2(x)$ where $P_1, P_2$ are exponential probability density functions with parameters $\lambda_1,\lambda_2$ respectively, then:
I : $\lambda_1\lambda_2=\lambda_1+\lambda_2$ is required for this to be a valid probability density function.
$$\int_{\Bbb R^+} P_X(x)\operatorname d x ~=~2-\dfrac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}$$
$$\lambda_2= \frac{\lambda_1}{\lambda_1-1}, \quad \lambda_1 > 1$$
II : The Expectation is:
$$\begin{align}\mathsf E(X) ~=~& \int_{\Bbb R^+} x~P(X=x)~\mathrm d~ x \\[1ex] =~& \int_{\Bbb R^+} x(P_1(x)+P_2(x)-P_1(x)P_2(x))~\mathrm d~ x\\[1ex]~ = ~& \frac 1{\lambda_1}+\frac 1{\lambda_2}-\int_{\Bbb R^+} x\lambda_1\lambda_2e^{-(\lambda_1+\lambda_2)x}~\mathrm d~ x \\[1ex] ~=~ & \frac 1{\lambda_1}+\frac 1{\lambda_2}-\frac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2)^2} \\[1ex] =~& 1-\lambda_1^{-1}+\lambda_1^{-2}\end{align}$$