Question 74150
an olymp[ic archer is able to hit the bullseye 80% of the time. assume each shot is independent of the othjers. if she shoots 6 arrows, whats the probability of the result below?
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P(hit)=0.8
P(miss)=0.2 
a) the 1st bullseye comes on the 3rd arrow
He has to miss,miss,hit
P(mmh)=(0.2)^2(0.8) =0.032
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b)she misses the bullseye at least once
P(at least one hit) = 1 - P(all misses)
= 1 - 0.2^6= 1-(6,4 x 10^-5)= 0.999936....
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c)the 1st bullseye comes on the 4th or 5th arrow
These are disjoint events, so:
P(4th or 5th) = 0.2^3*0.8+0.2^4*0.8 = 0.00768
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d)gets exactly 4 bullseyes
This is binomial with n=6,p=.8,x=4
If you have a TI calculator use binompdf(6,0.8,4)=0.24576
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e)gets at least 4 bullseyes
Use [1 - binomcdf(6,0.8,3)]=0.90112...
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f)gets at most 4 bullseyes
Use binomcdf(6,0.8,4)=0.34464
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Cheers,
Stan H.