Suppose there are $3$ red balls and $1$ blue ball. What's the probability that if I draw a ball $10$ times, I don't draw a blue ball?
Is it $1-\Pr(\text{no red ball}) = 1-\big(\frac{3}{4}\big)^3$?
Also, what is the chance that I draw at least $1$ blue ball within the $10$ draws?