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

d

step-by-step explanation:

d

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: