The surface of a circular dart board has a small center circle called the bull\'

Question

The surface of a circular dart board has a small center circle called the bull's-eye and 20 pie-shaped regions numbered from 1 to 20. Each of the pie-shaped regions is further divided into three parts such that a person throwing a dart that lands in a specific region scores that value of the number, double the number, or triple the number, depending on which of the three parts the dart hits. If a person hits the bull's-eye with probability 0.01, hits a double with probability 0.15, hits a triple with probability 0.20, and misses th

Explanation / Answer

let A be the number of bull's eyes in 8 throws

B be the number of triples in 8 throws

C be the number of doubles in 8 throws

D be the number of complete misses in 8 throws.

so total number of throws=8

probability of hitting a bull's eye=p1=0.01

probability of hitting a triple=p2=0.20

probability of hitting a double=p3=0.15

probability of complete miss=p4=0.64

hence (A,B,C,D) follows a multinomial distribution with parameter space (n,p1,p2,p3,p4)

with n=8, p1=0.01 p2=0.20   p3=0.15   p4=0.64

hence the probability mass function of (A,B,C,D) is

P[A=a,B=b,C=c,D=d]=[8!/(a!*b!*c!*d!)]*(0.01)a*(0.20)b*(0.15)c*(0.64)d   such that a+b+c+d=8

by question a=0 b=3 c=2 d=3

hence the probability of 0 bull's eyes, 3 triples , 2 doubles , 3 complete misses is

P[A=0,B=3,C=2,D=3]=[8!/(0!*3!*2!*3!)]*(0.01)0*(0.20)3*(0.15)2*(0.64)3

                               =560*(0.20)3*(0.15)2*(0.64)3=560*0.008*0.0225*0.262144=0.026424115 [answer]

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