We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. Using the result $H_n \sim \ln n$ as $n \to +\infty$, show that $$\mathbf { P } \left( A _ { n } \cup B _ { n } \right) \sim \frac { \ln n } { n } \quad ( n \rightarrow + \infty )$$

Let $Z$ be a random variable taking values in $\{ 0 , \ldots , M \}$. Show that:
$$\mathbf { E } [ Z ] = \sum _ { ( s , i , r ) \in E } \left( \sum _ { k = 0 } ^ { M } k \mathbf { P } \left( Z = k \mid \left( \tilde { S } _ { n } , \tilde { I } _ { n } , \tilde { R } _ { n } \right) = ( s , i , r ) \right) \right) \mathbf { P } \left( \left( \tilde { S } _ { n } , \tilde { I } _ { n } , \tilde { R } _ { n } \right) = ( s , i , r ) \right).$$

29. If $\vec { E }$ and $\vec { F }$ are the complementary events of events $E$ and $F$ respectively and if $0 < \mathrm { P } ( \mathrm { F } ) < 1$, then.
(A) $\quad P ( E / F ) + P ( \vec { E } / F ) = 1$
(B) $\quad P ( E / F ) + P ( E / \vec { F } ) = 1$
(C) $\mathrm { P } ( \vec { E } / \mathrm { F } ) + \mathrm { P } ( \mathrm { E } / \vec { F } ) = 1$
(D) $\quad \mathrm { P } ( \mathrm { E } / \vec { F } ) + \mathrm { P } ( \vec { E } / \vec { F } ) = 1$

Let $E^c$ denote the complement of an event $E$. Let $E$, $F$, $G$ be pairwise independent events with $P(G) > 0$ and $P(E \cap F \cap G) = 0$. Then $P(E^c \cap F^c | G)$ equals
(A) $P(E^c) + P(F^c)$
(B) $P(E^c) - P(F^c)$
(C) $P(E^c) - P(F)$
(D) $P(E) - P(F^c)$