SOLUTION: A circular archery target is 122 cm in diameter and consists of a bull's eye of diameter 12.2 cm, surrounded by nine evenly spaced concentric circles. Assume that an arrow that hit

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Question 255276: A circular archery target is 122 cm in diameter and consists of a bull's eye of diameter 12.2 cm, surrounded by nine evenly spaced concentric circles. Assume that an arrow that hits a target is equally likely to land anywhere on the target. What's the probability that the arrow will A) hit the bulls eye B) Land somewhere in the outmost four circles. C) Land within the fifth ring in from the outer edge. ---Please help on this problem! THANKS!
Found 2 solutions by drk, Theo:
Answer by drk(1908) About Me  (Show Source):
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Let's call the bulls eye - circle 1 including circle #1, there are 10 concentric circles.
radius of circle 1 = 6.1
radius of circle 2 = 12.2 or 6.1 * 2.
and so on.
A) hit the bulls eye
P(bulls eye) = pi*6.1^2 = 37.21pi cm^2
B) Land somewhere in the outmost four circles.
P(outer 4 circles) = p(whole target - inner 6 circles) = pi*R^2 - pi*r^2 = pi*61^2 -pi*36.6^2.
The 36.6 is the radius of the first 6 circles or 6.1 * 6.
This gives us 2381.44pi cm^2
C) Land within the fifth ring in from the outer edge.
P(5th ring) = p(5th ring - p(4th ring). We want only the fifth ring so all other inner rings must be subtracted out. We get pi*R^2 - pi*r^2 = pi*30.5^2 -pi*24.4^2 = 334.89pi cm^2

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
area of the target is equal to pi * r^2

radius of the target equals 122/2 = 61 cm.

radius of the bullseye = 12.2/2 = 6.1 cm.

outside of the bullseye there are 9 evenly spaced concentric rings.

this means that the width of each of those rings would equal (61 - 6.1)/9 = 6.1 cm each.

the target forms 10 concentric circles.

the bullseye is one of these circles and there are an additional 9 circles surrounding the bullseye.

the area of each of these circles is from the center of the target to the line forming that circle.

the circle formed by the bullseye has a radius of 6.1 cm.
the circle formed by the 1st ring has a radius of 12.2 cm.
the circle formed by the 2d ring has a radius of 18.3 cm.
the circle formed by the 3d ring has a radius of 24.4 cm.
the circle formed by the 4th ring has a radius of 30.5 cm.
the circle formed by the 5th ring has a radius of 36.6 cm.
the circle formed by the 6th ring has a radius of 42.7 cm.
the circle formed by the 7th ring has a radius of 48.8 cm.
the circle formed by the 8th ring has a radius of 54.9 cm.
the circle formed by the 9th ring has a radius of 61 cm.

the total area of the target is equal to pi * 61^2 = 11689.86626 cm^2.

this is the same as the area of the circle formed by the 9th ring.

the probability of hitting anywhere on the target is equal to 1.0 (100%). the assumption is that you will never miss the target.

the area of the bullseye = pi * 6.1^2 = 116.8986626

the probability of hitting the bullseye = 116.8986626 / 11689.86626 = .01

the probability of landing in the outmost 4 rings is calculated as follows:

4 outer rings are the 6th, 7th, 8th, and 9th rings.

the radius of the circle formed by the 9th ring is 61 cm.

the radius of the circle formed by the 5th ring is 36.6 cm.

the area from the center of the circle to a circle with a radius of 36.6 cm = pi * (36.6)^2 = 4208.251855 cm^2.

the area from the center of the circle to a circle with a radius of 61 cm = pi * 61^2 = 11689.86626 cm^2 (this is the area of the target we calculated earlier).

the area for the outer 4 circles is equal to the area of the circle formed by the 9th ring minus the area of the circle formed by the 5th ring.

the area for the outer 4 circles is therefore equal to:

11689.86626 - 4208.251855 = 7481.514409 cm^2.

the probability of hitting this area is equal to the area of the 4 outer rings divided by the total area of the target = 7481.514409 / 11689.86626 = .64

the area of the 5th ring only is the area of the circle formed by the 5th ring minus the area of the circle formed by the 4th ring.

that would be pi * (36.6)^cm^2 minus pi * (30.5)^2.

that equals 4208.351855 minus 2922.466566 = 1285.885289 cm^2

the probability of hitting this area would be 1258.885289 / 11689.86626 = .11