Question

The probability that an archer hits the 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 probability she hit the target

Answer 1

Case Where is a gust of wind and she hits the target

Let A denote the event that she hits the target on any given try and B be the event that there is a gust of wind on a given try. We are given

P(A|B) = 0.4, P(A|B') = 0.7 and P(B) = 0.3

P(A∩B) =P(B)P(A|B) = (0.3)(0.4) = 0.12

Answer 2

Answer:

0.61

Step-by-step explanation:

P(E) = Probability of Event E

Here, the event E is the even of the girl hitting the target

P(E) = P(E/A1)(PA1) + P(E/A2)*(P/A2) - (i)

Here, we have two situations, either the wind is flowing when she takes her shot or the wind is not flowing.

The even when wind flows can be considered as A1 and the other case A2.

Clearly, P(A1) = 0.3 and P(A2) = 1 - 0.3 = 0.7

Since, we know that when the wind is flowing, the probability of her hitting the target is 0.4, therefore we know P(E/A1) = 0.4, similarly, P(E/A2) = 0.7

Applying (i) formula we get

P(E) = 0.4*0.3 + 0.7*0.7 = 0.61