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

It is given that 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 of 1/32. It is assumed that the archer shoots 96 arrows at the target, and that all shoots are independent. We have to find the approximated probability that an archer hit not more than one bull. So, we have to find . We can see that, p = 1/32 and n = 96.

Let X be the number of bulls that the archer hits. We know that the binomial mass function p(x) =

Thus, = P(X=0) + P(X=1)

So, P(X = 0) =

P(X = 1) =

Thus, =  0.04746 + 0.14697 = 0.19443

Thus, = 0.19443

Now, Poisson approximation to the Binomial,

n = 96 and p = 1/32

Thus, = np = 96(1/32) = 3

We know,

Thus,

Thus, Thus,

Thus, this value is close to the exact value we have calculated above.