Question

Using the figure of the bullseye above, what is probability that a shooter will hit the 3rd ring in his first attempt and the bullseye on his second attempt? Express your answer as a decimal rounded to the nearest hundredths place.

0.7
0.07
0.53
0.17

Answer 1

So the answer is 0.07

Step-by-step explanation:

Area of the bullseye = 6^2 pi = 36 pi

Area of the white ring = (6 + 5)^2 pi - 6^2 pi = 85 pi

Area of outer blue ring = (6 + 5 + 5)^2 pi - (6 + 5)^2 pi = 135 pi

Total area of target = (6 + 5 + 5)^2 pi = 256 pi

If the shooter's throws are independent of one another, then

P(3rd ring AND bullseye) = P(3rd ring) * P(bullseye)

P(3rd ring) = (135 pi)/(256 pi) = 135/256

P(bullseye) = (36 pi)/(256 pi) = 9/64

P(3rd ring AND bullseye) = (135/256)*(9/64) = 1215/16384 = 0.07

Answer 2

Area of the bullseye = 6^2 pi = 36 pi

Area of the white ring = (6 + 5)^2 pi - 6^2 pi = 85 pi

Area of outer blue ring = (6 + 5 + 5)^2 pi - (6 + 5)^2 pi = 135 pi

Total area of target = (6 + 5 + 5)^2 pi = 256 pi

If the shooter's throws are independent of one another, then

P(3rd ring AND bullseye) = P(3rd ring) * P(bullseye)

P(3rd ring) = (135 pi)/(256 pi) = 135/256

P(bullseye) = (36 pi)/(256 pi) = 9/64

P(3rd ring AND bullseye) = (135/256)*(9/64) = 1215/16384 = 0.07