The problem related to Exponential distribution. Question:
There are two batteries. A Battery's life is following the Exp(1/20) distribution. The other one's life is following the Exp(1/40) distribution.
One day, a person randomly chose one battery. When the battery was used for 20 hours, find the probability that the battery can be alive 10 more hours.

Answer from solution:
$A_1: \mbox{the case of choosing 1st batery}$
$A_2: \mbox{the case of choosing 2st batery}$
$X: \mbox{The duration time of battery}$
$X|_{A_1} \sim Exp(1/20)$
$X|_{A_2} \sim Exp(1/40)$
\begin{align}
P(X>30~|~X>20)=\frac{P(X>30)}{P(X>20)}&=\frac{P(A_1)P(X>30~|~A_1)+P(A_2)P(X>30~|~A_2)}{P(A_1)P(X>20~|~A_1)+P(A_2)P(X>20~|~A_2)}\\
&=\frac{\displaystyle\frac12\cdot e^{-\frac{30}{20}} + \frac12\cdot e^{-\frac{30}{40}}} {\displaystyle\frac12\cdot e^{-\frac{20}{20}} + \frac12\cdot e^{-\frac{20}{40}}}\\
&=\frac{e^{-\frac{30}{20}} + e^{-\frac{30}{40}}} {e^{-\frac{20}{20}} + e^{-\frac{20}{40}}}
\end{align}

Answer from me:
\begin{align}
P(X>30~|~X>20) &= P(A_1)~P((X>30~|~X>20)~|~A_1) + P(A_2)~P((X>30~|~X>20)~|~A_2)\\
&=P(A_1)~P(X>10~|~A_1) + P(A_2)~P(X>10~|~A_2)\\
&~~~~~(\because \mbox{memory-less property of exponential distribution})\\
&= \frac12 e^{-\frac{10}{20}}+ \frac12 e^{-\frac{10}{40}}\\
&= \frac12 \left(e^{-\frac{1}{2}}+ e^{-\frac{1}{4}}\right)
\end{align}

I agree with the solving process of solution.
However, I don't understand where my solving process is wrong.
Please enlighten thick-headed-me.
 A: The problem with your solution is that you use the following partly undefined identity:
$$P(X>30\mid X>20) = P(A_1)P((X>30\mid X>20)\mid A_1) + P(A_2)P((X>30\mid X>20)\mid A_2).$$
Namely the definition of $P((X>30\mid X>20)\mid A_1)$ is not clear because $(X>30\mid X>20)$ is not defined as an event.
Let's jump over the undefined concept of the "double conditional" and let's just see your result: 
$$P(X>30\mid X>20)=P(X>10\mid A_1)P(A_1)+P(X>10\mid A_2 )P(A_2)$$
which you verify by referring to the memory-less nature of the exponential distribution. 
Unfortunately 
$$X\sim\begin{cases}
\exp(\lambda_1),& \text{ with probability } \frac12\\
\exp(\lambda_2),& \text{ with probability } \frac12
\end{cases}$$
is not exponentially distributed:
$$F_X(x)=P(X<x)=P(X<x\mid A_1)P(A_1)+P(X<x\mid A_2)P(A_2)=$$
$$=\left(2-e^{\lambda_1x}-e^{\lambda_1x}\right)\frac12.$$
And the density is
$$f_X(x)=\frac12\left(\lambda_1e^{\lambda_1x}+\lambda_2e^{\lambda_2x}\right).$$
This distribution is not memory-less.
