Question

An archer shoots arrows at a circular target where the central portion of the target inside is called the bull. The archer hits the bull with probability 1/32. Assume that the archer shoots 96 arrows at the target, and that all shoots are independent. What is an approximated probability that an archer hit not more than one bull?

Answer 1

This is a binomial distribution with $p=\frac{1}{32},\;n=96.$

$P(X\le1)=P(X=0)+P(X=1)=C_{96}^0(\frac{1}{32})^0(\frac{31}{32})^{96}+C_{96}^1(\frac{1}{32})^1(\frac{31}{32})^{95}=0.1944.$