Question

The probability that an archer hits her target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that a. on a given shot there is a gust of wind and she hits her target.
b. she hits the target with her first shot.

c. she hits the target exactly once in two shots.

d. there was no gust of wind on an occasion when she missed.

Answer 1

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