document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "a) the 1st bullseye comes on the 3rd arrow
\n" ); document.write( "b)she misses the bullseye at least once
\n" ); document.write( "c)the 1st bullseye comes on the 4th or 5th arrow
\n" ); document.write( "d)gets exactly 4 bullseyes
\n" ); document.write( "e)gets at least 4 bullseyes
\n" ); document.write( "f)gets at most 4 bullseyes
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Algebra.Com's Answer #53178 by stanbon(75887)\"\" \"About 
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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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\n" ); document.write( "P(hit)=0.8
\n" ); document.write( "P(miss)=0.2
\n" ); document.write( "a) the 1st bullseye comes on the 3rd arrow
\n" ); document.write( "He has to miss,miss,hit
\n" ); document.write( "P(mmh)=(0.2)^2(0.8) =0.032
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\n" ); document.write( "b)she misses the bullseye at least once
\n" ); document.write( "P(at least one hit) = 1 - P(all misses)
\n" ); document.write( "= 1 - 0.2^6= 1-(6,4 x 10^-5)= 0.999936....
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\n" ); document.write( "c)the 1st bullseye comes on the 4th or 5th arrow
\n" ); document.write( "These are disjoint events, so:
\n" ); document.write( "P(4th or 5th) = 0.2^3*0.8+0.2^4*0.8 = 0.00768
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\n" ); document.write( "d)gets exactly 4 bullseyes
\n" ); document.write( "This is binomial with n=6,p=.8,x=4
\n" ); document.write( "If you have a TI calculator use binompdf(6,0.8,4)=0.24576
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\n" ); document.write( "e)gets at least 4 bullseyes
\n" ); document.write( "Use [1 - binomcdf(6,0.8,3)]=0.90112...
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\n" ); document.write( "f)gets at most 4 bullseyes
\n" ); document.write( "Use binomcdf(6,0.8,4)=0.34464
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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