Probability and statistics expected value

Definitely a challenge...
A target has 3 concentric circles of 1, 3, and 5 inches, respectively. An archer scores 10 Pts for hitting a bullseye, 5 for hitting the middle, and 3 for hitting the outer region. If the archer hits the target with a probability if 1/2 and Is equally likely to hit one point on the target as any other, find the expected number E of points he scores each time he fires.

I'm interpreting this problem differently from one of the answerers: all locations (not point values) on the target are equally likely, so that that the probability of hitting a region is proportional to the area of the region (if the target is hit at all), so the sizes of the regions become relevant. This is a much more realistic interpretation.

The area of the bull's eye (worth 10 points) is pi(1^2) = pi square inches.
The area of the middle region (worth 5 points) is pi(3^2 - 1^2) = 8 pi square inches.
The area of the outer region (worth 3 points) is pi(5^2 - 3^2) = 16 pi square inches.
The area of the whole target is pi(5^2) = 25 pi square inches.

Since the probability of hitting the target is 1/2, and probability of hitting a region is proportional to the area of the region,

P(score 10 points) = (1/2)(pi / (25 pi)) = 1/50
P(score 5 points) = (1/2)(8 pi / (25 pi)) = 4/25
P(score 3 points) = (1/2)(16 pi / (25 pi)) = 8/25
P(score 0 points) = 1 - 1/2 = 1/2

Finally, the expectation of the score on each attempt is

10(1/50) + 5(4/25) + 3(8/25) + 0(1/2) = 0.2 + 0.8 + 0.96 + 0 = 1.96

Lord bless you today!

You see, there are four possible outcomes here. There's the possibility that the archer hits the inner circle, the middle circle, the outer circle, or misses. The probability that the archer misses is .5, or 1/2. The probability that the archer hits any of the circles is 1/6. Also, the inner circle is worth 10 points, middle is worth 5, and outer is worth 3 (obviously, missing gives him no points). So now, you do the math.