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The question:

Mike and John are playing a friendly game of darts where the dart board is a disk with radius of 10in. Whenever a dart falls within 1in of the center, 50 points are scored. If the point of impact is between 1 and 3in from the center, 30 points are scored, if it is at a distance of 3 to 5in 20 points are scored and if it is further that 5in, 10 points are scored. Assume that both players are skilled enough to be able to throw the dart within the boundaries of the board. Mike can place the dart uniformly on the board (i.e., the probability of the dart falling in a given region is proportional to its area). (a) What is the probability that Mike scores 50 points on one throw? (b) What is the probability of him scoring 30 points on one throw? (c) John is right handed and is twice more likely to throw in the right half of the board than in the left half. Across each half, the dart falls uniformly in that region. Answer the previous questions for John’s throw

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I have the answers I'm just not quite getting it.

For a, why is the formula απR2 = απ? Why is there a constant and pie on both sides?

For b specifically, why is the formula equal to απ32 = 0.09. Why is it 3 squared for more than 30 pts? As wouldn't 3 be within the 30pt category?

this is the answer for a: The probability of Mike scoring 50 points is proportional to the area of the inner disk. Hence, it is equal to απR2 = απ, where α is a constant to be determined. Since the probability of landing the dart on the board is equal to one, απ102 = 1, which implies that α = 1/(100π). Therefore, the probability that Mike scores 50 points is equal to π/(100π) = 0.01

this is the answer for b: In order to score exactly 30 points, Mike needs to place the dart between 1 and 3 inches from the origin. An easy way to compute this probability is to look first at that of scoring more than 30 points, which is equal to απ32 = 0.09. Next, since the 30 points ring is disjoint from the 50 points disc, probability of scoring more than 30 points is equal to the probability of scoring 50 points plus that of scoring exactly 30 points. Hence, the probability of Mike scoring exactly 30 points is equal to 0.09 − 0.01 = 0.08

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    $\begingroup$ I don't think it matters whether it falls on the left or right half. Both probabilities should be the same. $\endgroup$
    – mjw
    Commented Aug 10, 2021 at 0:57
  • $\begingroup$ "pie on both sides"? Good apetite. $\endgroup$
    – Jean Marie
    Commented Aug 11, 2021 at 17:13
  • $\begingroup$ why did you delete your question? math.stackexchange.com/questions/4228457/… $\endgroup$
    – BCLC
    Commented Aug 19, 2021 at 18:57
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    $\begingroup$ @BCLC I am re-posting it in an attempt to make it more clear but I have to wait 40-50 minutes. $\endgroup$
    – user946705
    Commented Aug 19, 2021 at 19:01
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    $\begingroup$ @BCLC I have edited it and undeleted it. I hope it is clear now. $\endgroup$
    – user946705
    Commented Aug 19, 2021 at 19:13

3 Answers 3

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The probability for both players are the same!

Player 1: P(50 points) = $\frac{1^2}{10^2} = \frac{1}{100}$, as you noted.

Player 2: P(50 points) = $\frac{2}{3} \frac{1^2/2}{10^2/2} + \frac{1}{3} \frac{1^2/2}{10^2/2} = \frac{1}{100}.$

Similarly for each other score.

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The probability of landing inside a circle of radius r is propotional to the area of that circle.

$$P(x<r) = C \pi r^2$$

To find that constant C, we can use the fact that $P(x<R)$ should be 1.

This implies that $C \pi R^2=1$ and that

$$C=\frac{1}{\pi R^2}$$

This way, we have found that,

$$P(x<r) = \frac{r^2}{R^2}$$

The probability that a dart falls within 1in of the center, is

$$P(x<1in) = \frac{(1 in)^2}{(10 in)^2}=\frac{1}{100}$$

As the probability of throwing to the right or the left of the left has no effect on the distance to the center, this dooes not influence the result.

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Short answer: as given in your problem: "(i.e., the probability of the dart falling in a given region is proportional to its area)." and the area of a disk is what now?


For a, why is the formula απR2 = απ? Why is there a constant and pie on both sides?

this is the answer for a: The probability of Mike scoring 50 points is proportional to the area of the inner disk. Hence, it is equal to απR2 = απ, where α is a constant to be determined. Since the probability of landing the dart on the board is equal to one, απ102 = 1, which implies that α = 1/(100π). Therefore, the probability that Mike scores 50 points is equal to π/(100π) = 0.01

Since the dart lands with uniform distribution over the area of the 10" board, then the probability that it falls in a smaller area is equal to the ratio of that area to the area of the board.   Both areas are disks, and the area of a disk equals $\pi$ times the square of the radius ($\pi r^2$), then the common factor ($\pi$) will cancel, and the ratio we seek is just the ratio of the square of the radii: namely $1^2/10^2$.

For b specifically, why is the formula equal to απ32 = 0.09. Why is it 3 squared for more than 30 pts? As wouldn't 3 be within the 30pt category?

The given answer is overly complicated. The favoured area is now an annulus, whose area is the difference between its outer disk and the excluded inner disk: $\pi(r_\mathrm o^2-r_\mathrm i^2)$, and here the relevant radii are $r_\mathrm i=1$ and $r_\mathrm o=3$.   Again, the common factor cancels, so the probability of landing within this favoured annulus is simply $(3^2-1^2)/10^2$.

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  • $\begingroup$ That doesn't add up to 0.09? $\endgroup$
    – user946705
    Commented Aug 10, 2021 at 17:56
  • $\begingroup$ Oh, never mind...0.08 to be in that area + 0.01 from the 1inch. $\endgroup$
    – user946705
    Commented Aug 10, 2021 at 17:59

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