0

My question is similar to the one here. However, is the probability of hitting closer to the center than the edge of a rectangular dartboard going to involve an almost identical process to that of a square? Using a square board of side 2 results in the integration of $f(x)=\frac{1-x^2}{2}$ and $g(x)=x$ so that $$\int_0^{\sqrt{2}-1}\frac{1-x^2}{2}-1\ dx$$ gives us an 1/8 of the space closer to the center rather than the edge, with a final probability coming to about 22% chance. Would using a rectangular dartboard involve the same calculations and equations of a square, using a square to calculate the are closer to the center and dividing by the total area of the rectangle? Any help solving this is greatly appreciated! Thanks.

4ntim4tter
  • 11

1 Answers1

1

Yes, the rectangle is solved just like the square. Indeed, if we have a rectangle with a shorter side 2, then any point which is outside the square with side 2 with the same center is further than 1 away from the center but no more than 1 away from some side. So the area which is closer to the center is the same as in square.

colt_browning
  • 3,647
  • Thanks for the speedy reply. Is there a way to easily generalize the same problem for a square of any side? – 4ntim4tter Dec 02 '17 at 21:57
  • 1
    Well, the portion of the area of the square that is closer to the center does not depend on the side length. So if you have an answer for a square of side 2, divide it by 4 and multiply by the square of the shortest side of the rectangle to get the area which is closer to the center of the rectangle, then divide by the area of the rectangle to get the probability. – colt_browning Dec 02 '17 at 22:55