Probability and statistics expected value?

Question

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.

Answer 1

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!

Answer 2

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.



It's actually quite simple, really. All you do is multiply each value by it's respective probability of outcome and add them all up. So it looks like this:

E=10(1/6) + 5(1/6) + 3(1/6) + 0(1/2)



So now you just do the math:

E=10(1/6) + 5(1/6) + 3(1/6) + 0(1/2)

E=10/6 + 5/6 + 3/6 + 0/6

E=18/6

E=3



So the expected number of points the archer is expected to make on any given shot is 3 points.











It would appear as though I have made a mistake. You said "...equally likely to hit one point on the target as any other...". This means that the areas must be taken into account for the probability. I could do this, but I don't care enough to, so just check the other answers. It appears as though they've got it right.

Answer 3

Notice that the target is 1/25 bullseye, 8/25 centre, 16/25 outer. (by area)



So:

1/2 * 1/25 = 1/50 of the time he scores 10

1/2 * 8/25 = 8/50 of the time he scores 5

1/2 * 16/25 = 16/50 of the time he scores 3

1/2 = 25/50 of the time he scores 0



Then E(x) = sum x p(x)

= 10 * 1/50 + 5 * 8/50 + 3 * 16/50 + 0 * 25/50



EDIT: Removed a stray "* 5" from the last line. Total of 98/50 = 1.96 agrees with Ian.

Answer 4

If the the archer has a probability of 1/2 of hitting the target at all, then half the time the archer will receive 0 points and the other half of the time they will receive a positive amount of points. Since it is mentioned that one point value on the target is equally likely to another point value, then we can ignore the area of the each point value and deviate the remaining 1/2 probability equally between the point values. So, you should have determined that:



Points---------------Probability

0---------------------------1/2

3---------------------------1/6

5---------------------------1/6

10-------------------------1/6



Therefore, the expected value of points on a single throw would be



3(1/6)+5(1/6)+10(1/6)

= 3 points <--------------------Answer

Answer 5

What Class is this for