Let $X_1 \sim \boldsymbol\exp (\lambda_1)$ and $X_2 \sim \boldsymbol\exp (\lambda_2)$. Then, $$\mathbb{P} \{X_1 < X_2 | X_1 = x\} = \mathbb{P}\{x < X_2 \}.$$ (Ross Introduction to Probability Models page 302)
Yet, if I use the definition of the conditional expectation $$\mathbb{P}\{A|B\} = \mathbb{P}\{A \cap B\} / \mathbb{P}\{A\},$$ we get devision by zero, because the probability $\mathbb{P}\{X_1 = x\}$ is zero since $X_1$ follows a continous distribution. What am I missing here?
The exponential distribution is irrelevant. The independence assumption, which is relevant, is missing.
The following fact follows from Fubini.
In our case, for any measurable $A$, $$ \mathrm{E} [\mathbf{1}_{X_1<X_2} \mathbf{1}_{X_1\in A}] = \mathrm{E} [\mathrm{E} [\mathbf{1}_{x<X_2}]|_{x=X_1} \mathbf{1}_{X_1\in A}] = \mathrm{E} [\mathrm{P}(x<X_2)|_{x=X_1} \mathbf{1}_{X_1\in A}], $$ whence by definition $$ \mathrm{P} [{X_1<X_2}\mid X_1 = x] = \mathrm{E} [\mathbf{1}_{X_1<X_2}\mid X_1 = x] = \mathrm{P}(x<X_2). $$