Hans and Franz are in a shooting competition. The object of the match is to be the first to hit the bulls-eye of a target 100 feet away. The two opponents alternate turns shooting, and each opponent has a 40% chance of hitting the bulls-eye on a given shot. If Hans graciously allows Franz to shoot first, what is the probability that Hans will win the competition and take no more than three shots?

Question

Victor Mccall

Mathematics

4 January, 05:06

Hans and Franz are in a shooting competition. The object of the match is to be the first to hit the bulls-eye of a target 100 feet away. The two opponents alternate turns shooting, and each opponent has a 40% chance of hitting the bulls-eye on a given shot. If Hans graciously allows Franz to shoot first, what is the probability that Hans will win the competition and take no more than three shots?

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Answers (1)

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Milo Wong · Answer 1

Step-by-step explanation:

P (success) = 0.4 for both Hans and franz

P (failure) = 0.6, n = 100

what is the probability that Hans will win the competition and take no more than three shots;

P (x < 3) = P (x=0) + P (x = 1) + P (x = 2)

From binomial probability; nCx P^x q^ (n-r)

= 100C0 x 0.4^0 + 0.6^100 + 100C1 X 0.4^1 X 0.6^99 + 100C2 X 0.4^2 + 0.6^98

= 1.48 x 10^-19