I throw darts repeatedly. Assume that on each single throw, my chance of hitting the bulls-eye is $10\%$, independently of all other throws. I decide to throw until I have hit the bulls-eye $3$ times. What is the chance that I throw exactly $30$ times?
$P(\text{throw exactly }30\text{ times})= (1-0.10)^{30}=0.04239=4.239\%$
Best Answer
The third success occurs on the $30$th trial if and only if there are exactly $2$ successes in the first $29$ trials, and then a success on the $30$th.
If $p$ is the probability of success on any trial, here $0.1$, then the probability of $2$ successes in $29$ trials is $\binom{29}{2}p^2(1-p)^{27}$. Now multiply by $p$.