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.

Step-by-step explanation:

Let T be the event that she hot the target and W be the event that there was a gust of wind.

P(T | W) = 0.4

P(T | W') = 0.7

P(W) = 0.3

P(W') = 1 - 0.3 = 0.7

a) P(W T) = P(W) x P(T | W)

= 0.3 x 0.4

= 0.12

b) P(T) = P(W T) + P(W' T)

= 0.12 + 0.7 x 0.7

= 0.61

c) P(she hits the target exactly once in two shots) = P(hits the target on first shot and misses on the second) + P(misses the target on first shot and hits on the second)

= 0.61 x (1-0.61) + (1-0.61) x 0.61

= 0.4758

d) P(W' | T') = P(T'W')/P(T') [Bayes' theorem]

= [0.7 x (1-0.7)]/[0.7 x (1-0.7) + 0.3 x (1-0.4)]

= 0.5385

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation:

22

step-by-step explanation: