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?

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

bfgbh bhf gfdzf bhgfrdfgbn cvdgsytrghvn ncvhgfdtrseawa hgdrsxcvbgfrewq

step-by-step explanation:

plz mark brainliest i took the test