Throw an arrow $\rightarrow$

For 10 independent reoccuring events (sampling with replacement) with the same probability for each individual event, we use binomial theorem. So with probability of and the event to happen is 0.9, and we want 9 out of 10 targets, we can expand the binomial to the $10^{th}$ power and find the coefficient of exactly 9 successes.

Namely let $p=(0.9a+0.1)$ where a is a label for a success. Thus $(0.9a+0.1)^{10}$ is the probability of 10 events.

Let $P(x=k) \sim Binom(n=10,\pi = 0.9) \\ =\binom{10}{k}(0.9^{k})(0.1^{10-k})$

Thus the coefficient for 9 successes is as follows.

$P(x=9)=\binom{10}{9}(0.9^{9})(0.1^{1})\\=\frac{10!}{9!\cdot 1!}(0.9^{9})(0.1)\\=\frac{10\cdot 9!}{9!}(0.9^{9})(0.1)\\=10(0.9^{9})(0.1) \\ = 0.9^{9}\\ = 0.38742\\ \approx 38.74 \%$