Two archers, A and B, are shooting at the same target. A hits the target 75% of the time and B hits the target 25% of the time. Now suppose that both archers shoot one arrow at the target at the same time.
If exactly one arrow hits the target, what is the probability that it was shot by A?

P ( A wins / only 1 hits ) = 0.9

Step-by-step explanation:

Given:-

- Probability to hit target by Archer A, pa = 0.75

- Probability to hit target by Archer B, pb = 0.25

Find:-

f exactly one arrow hits the target, what is the probability that it was shot by A?

Solution:-

- We are required to compute the conditional probability such that Archer A hit the target given that only one arrow hits.

           P ( A wins / only 1 hits ) = P ( A win & 1 hit ) / P ( 1 hit )

- The probability if the target is only hit by Archer A is that A hits and B misses since the success of hitting the target is independent for both archers we can write:

          P ( A wins & 1 hit ) = pa*(1-pb)

                                        = (0.75)*(1 - 0.25) = 0.75^2

                                        = 0.5625

- The probability of only one of the archer hits the target will constitute of two cases. Case I: A hits and B misses ; Case II : A misses and B hits.

         P ( Only 1 hits ) = pa*(1-pb) + (1-pa)*pb

                                  =(0.75)*(1 - 0.25) + (1 - 0.75)*(0.25)

                                  = 0.75^2 + 0.25^2

                                  = 0.625

- Plug in the respective probabilities in the conditional probability expression:

        P ( A wins / only 1 hits ) = 0.5625 / 0.625

                                                = 0.9

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

step-by-step explanation: