exercise08.03.tex

\paragraph{Exercise 8.3} Let $X$ be a uniform random variable on [0,1]. Then,
\[
    \pr\left(X \leq \frac{1}{2} \mid \frac{1}{4} \leq X \leq \frac{3}{4}\right)
    = \frac{\pr\left(\frac{1}{4} \leq X \leq \frac{1}{2}\right)}{\pr\left(\frac{1}{4} \leq X \leq \frac{3}{4}\right)}
    = \frac{\frac{1}{2} - \frac{1}{4}}{\frac{3}{4} - \frac{1}{4}}
    = \frac{1}{2},
\]
and
\begin{align*}
    \pr\left(X \leq \frac{1}{4} \mid \left(X \leq \frac{1}{3}\right) \cup \left( X \geq \frac{2}{3}\right) \right)
    &= \frac{\pr\left(\left(X \leq \frac{1}{4} \right) \cap \left( \left(X \leq \frac{1}{3}\right) \cup \left( X \geq \frac{2}{3}\right) \right) \right)}{\pr\left(\left(X \leq \frac{1}{3}\right) \cup \left( X \geq \frac{2}{3}\right) \right)} \\
    &= \frac{\pr\left(X \leq \frac{1}{4}\right)}{\pr\left(X \leq \frac{1}{3} \right) + \pr\left(X \geq \frac{2}{3}\right)} \\
    & = \frac{\frac{1}{4}}{\frac{1}{3} + \frac{1}{3}} \\
    &= \frac{3}{8}.
\end{align*}