Question

Solve using R - An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what’s the probability of each of the following results? a) Her first bull’s-eye comes on the third arrow. b) She misses the bull’s-eye at least once. c) Her first bull’s-eye comes on the fourth or fifth arrow. d) She gets exactly 4 bull’s-eyes. e) She gets at least 4 bull’s-eyes. f) She gets at most 4 bull’s-eyes.

Answer 1

a)

P( Her first bull’s-eye comes on the third arrow)=P(first 2 arrow does not hit and 3rd arrow hits)

=(1-0.8)²*0.8 =0.0320

b)

P(She misses the bull’s-eye at least once)=1-P(hit all 6 times) =1-(0.8)⁶ =1-0.262144 =0.737856

c)

P(Her first bull’s-eye comes on the fourth or fifth arrow)) =P(first 3 arrow does not hit and 4th arrow hits)+P(first 4 arrow does not hit and 5th arrow hits) =(1-0.8)³*0.8+(1-0.8)⁴*0.8=0.00768

d)

P(She gets exactly 4 bull’s-eyes) =(6C4)*(0.8)⁴(0.2)² =0.24576

e)

P( She gets at least 4 bull’s-eyes) =P(X>=4)=P(X=4)+P(X=5)+P(X=6)

=(6C4)*(0.8)⁴(0.2)² +(6C5)*(0.8)⁵(0.2)¹ +(6C6)*(0.8)⁶(0.2)⁰ =0.90112

f)

P(She gets at most 4 bull’s-eyes) =P(X<=4)=1-P(X>=5)=1-P(X=5)-P(X=6)

=1-(6C5)*(0.8)⁵(0.2)¹ -(6C6)*(0.8)⁶(0.2)⁰ =1-0.65536 =0.34464