Question #115122

B3. 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

Expert's answer


Let X=X= the number of bulls that the archer hits: X∼Bin(n,p)X\sim Bin(n,p)

The binomial mass function


p(x)=P(X=x)=(nx)px(1−p)n−xp(x)=P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}

Given p=1/32,n=96p=1/32,n=96

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


p(x)=P(X=x)=(96x)(132)x(1−132)96−xp(x)=P(X=x)=\binom{96}{x}\bigg({1\over 32}\bigg)^x\bigg(1-{1\over 32}\bigg)^{96-x}

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


P(X≤1)=P(X=0)+P(X=1)=P(X\leq1)=P(X=0)+P(X=1)=

=(960)(132)0(1−132)96−0+=\binom{96}{0}\bigg({1\over 32}\bigg)^0\bigg(1-{1\over 32}\bigg)^{96-0}+


+(961)(132)1(1−132)96−1=+\binom{96}{1}\bigg({1\over 32}\bigg)^1\bigg(1-{1\over 32}\bigg)^{96-1}=


=(3132)95(12732)≈0.19443154=\bigg({31\over 32}\bigg)^{95}\bigg({127\over 32}\bigg)\approx0.19443154

Poisson Approximation to the Binomial


n=96>50>20,p=132=0.03125,np=3<5n=96>50>20, p={1\over 32}=0.03125, np=3<5

λ=np=3\lambda=np=3

P(X=0)≈e−3300!=e−3P(X=0)\approx{e^{-3}3^0\over 0!}=e^{-3}

P(X=1)≈e−3311!=3e−3P(X=1)\approx{e^{-3}3^1\over 1!}=3e^{-3}

P(X≤1)=P(X=0)+P(X=1)≈P(X\leq1)=P(X=0)+P(X=1)\approx

≈e−3+3e−3=4e−3≈0.19914827\approx e^{-3}+3e^{-3}=4e^{-3}\approx0.19914827


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