$P[X_1Given $X_1$ and $X_2$ are exponentially distributed with parameter $\lambda_1$ and $\lambda_2$ respectively, we can see 
$P[X_1 < X_2] = \int_0^{\infty}P[X_2 > x | X_1 = x]P[X_1=x]=\frac{\lambda_1}{\lambda_1+\lambda_2}$. 
Given the simplicity of this result is there a way to prove this more intuitively like just from the first principles that 
$P[X_i \in (x,x+dx)] = \lambda_idx, i=1,2$ 
 A: My acknowledgements to André for his discussions that helped me figure out this answer. Let $A$ denote the event $min(X_1, X_2) \in (x,x+dx)$. 
\begin{eqnarray*}
P(X_1 < X_2) &=& \int_0^{\infty}\left[P(X_1 < X_2 | A)P(A)\right] \\
&=& \int_0^{\infty}\left[\frac{\lambda_1dx}{\lambda_1dx+\lambda_2dx}P(A)\right] \\
&=& \frac{\lambda_1}{\lambda_1+\lambda_2}\int_0^{\infty}P(A) \\
&=& \frac{\lambda_1}{\lambda_1+\lambda_2}
\end{eqnarray*}
Similar arguments show that, given two coins with probabilities $p_1$ and $p_2$ for head, the probability that the first tail happens simultaneously in a series of infinite tosses is $\frac{(1-p_1)(1-p_2)}{1-p_1p_2}$ without needing to calculate the infinite sums. 
A: the best way to solve this is like this
$$\begin{align}P(X_1<X_2)
&=\int^\infty_0P(X_1<X_2|X_1=x)f(x)dx\\
&=\int^\infty_0P(x<X_2|X_1=x)f(x)dx \\
&=\int^\infty_0P(X_2>x|X_1=x)f(x)dx\end{align}$$
because $X_2>x$ and $X_1=x$ are independent, so that
$$\begin{align}P(X_1<X_2)
&=\int^\infty_0P(X_2>x)f(x)dx\\
&=\int^\infty_0(1-1+e^{-\lambda_2x})\lambda_1e^{-\lambda_1x}dx\\
&=\int^\infty_0\lambda_1e^{-\lambda_1x-\lambda_2x}dx\\
&=\frac{\lambda_1}{\lambda_1+\lambda_2}\\
\end{align}$$
