Answer:
a) the probability is 0.12 (12%)
b) the probability is 0.61 (61%)
c) the probability is 0.476 (47.6%)
a) the probability is 0.538 (53.8%)
Step-by-step explanation:
a) denoting the event H= hits her target and G= a gust of wind appears hen
P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)
b) for any given shot
P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)
c) denoting P₂ as the probability of hitting once in 2 shots and since the archer can hit in the first shot or the second , then
P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)
d) for conditional probability we can use the theorem of Bayes , where
M= the archer misses the shot → P(M) = 1- P(H) = 0.39
S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7
then
P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)
where P(S/M) is the probability that there was no wind when the archer missed the shot
