= Solution
Assume that wrong-note events are <independent events> and that each note has the same error <probability> $p$, with $0\leq p\leq1$. Then their count $N$ has a <binomial distribution>. Choose the $n$ mistaken positions in $\binom{100}{n}$ ways and multiply the probabilities of the errors and correct notes:
$$
\boxed{\mathbb P(N=n)=\binom{100}{n}p^n(1-p)^{100-n},\qquad n=0,1,\ldots,100.}
$$
The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, so
$$
\boxed{\mathbb P(\text{booing})=\sum_{n=51}^{100}\binom{100}{n}p^n(1-p)^{100-n}.}
$$
The single-note error <probability> alone would not determine this count distribution if mistakes were correlated; <independence> is the additional standard modeling assumption.
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