Statistics and probability

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

(a) Find the probability mass function of the number of bulls that the archer hits.

(b) Give an approximation for the probability of the archer hitting no more than one bull

Answer 1

a P(x) = C(96,x)(1/32^x)(31/32 ^ (96-x)) or binom.dist(x,96,1/32,0)

b) Using a), P(0) + P(1) = C(96,0)(1/32^0)(31/32 ^96) + C(96,1)(1/32^1)(31/32 ^ 95) =

31/32 ^ 96 + 96 * 1/32 * 31/32 ^ 95 = 0.194431540212148 or, using a package like Excel, =BINOM.DIST(1,96,1/32,1) gives the same answer.

In the "old" days, one approximated the binomial in cases such as this with the Poisson distribution, and we would use np = 1/32 * 96 = 3 for the rate.

Then, P(x) = e^-3 3^x/x!

P(0) = e^-3; P(1) = 3 e^-3; P(0) + P(1) = 4e^-3 = 0.199148273471456; this value is "close" to the exact value above.