Question

Using the figure of the bullseye above, what is the probability that a shooter will hit the 3rd ring? Express your answer as a percentage rounded to the nearest whole percent.

Answer 1

So the answer is 0.07

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

The probability of hitting the third ring is the same as finding what percentage is the area of the third ring out of the total area of the board.

The area of the third ring = The total area of the board - The area of the second circle.

Refer to the diagram below
The diameter for the whole circle = 32 in
The radius = 16 in
The area of the whole circle = π(16)² = 256π

The diameter for the second circle = 22 in
The radius = 11 in
The area of the second circle = π(11)² = 121π

The area of the third ring = 256π - 121π = 135π

Area of the third ring as a percentage of the total area =
`(135 \pi )/(256 \pi ) *100` = 52.7%