A skilled sharpshooter misses a mark $4$ percent of the time. Find the probabilities
$(1)$ He will miss at most $1$ mark in $15$ shots.
$(2)$ He will miss the mark for the first time on the sixth shot.
Binomial Distribution :
$$\sum^{1}_{x=0} b(x,15,,04) = \bigl(\begin{smallmatrix} 15 \\0\end{smallmatrix}\bigr)\cdot(.4)^0(1-.04)^{15-0}+ \bigl(\begin{smallmatrix} 15 \\1\end{smallmatrix}\bigr) \cdot(.4)^1(1-.04)^{15-1} $$
$$ \Rightarrow\sum^{1}_{x=0} b(x,15,,04) = .542+3.3880 = 3.9301$$
However the answer according to the book is $.8809$ ? Does anyone know where I might have gone wrong?
Since the first question seems to have been answered by the comment of nicola, I'll answer the second, which calls for the geometric distribution, since order matters.
With $p=0.04$, we have $$(1-p)^5p=0.0326\dots$$