Answer to Question #119395 in Statistics and Probability for Michael

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

Here it is given that probability of hitting the bull is 1/32

And all events are independent.

And he shots 96 arrows at the bull and we need to determine the probability such that archer hit not more than one bull.

Probability = P(shooting 0 times out of 96) + P(shooting 1 time out of 96) = $(1- \frac{1}{32})^{96} + 96 \times (1- \frac{1}{32})^{95} \frac{1}{32} = 0.1944315$