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Let's say the problem is

In a game, you throw a circular coin with radius 1 unit onto a square board having side 10 units. A circle with radius 2 units is shaded in the center of the board. You win the game if the circle is able to touch the interior of the circle or at least its boundary. What is the probability that you win the game? Note: The coin will always land completely within the board.

Obviously, the solution would be much more simple if the coin is replaced with a dart, which is treated as if it lands on a single point.

How do you solve this now that you have to take into account the area of the coin too?

SolvingTrainee
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3 Answers3

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If the coin will always land completely within the board, then the center of the coin will always have to be at a minimum distance of 1 unit from the edges of the board. The sample space is therefore $8^2 = 64$. If the coin has to touch the circle inscribed at the center, it's center should fall within the circle with radius 3 drawn at the center of the square. The event space is hence $9\pi$. The probability that you win the game is $9\pi/64$.

Kaushik
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Hint: think about where the center of the coin needs to land for its edge to land somewhere inside of the circle. This is an area larger than the circle, since the coin has a positive area. Then you can treat the coin like its a dart, as you mentioned.

Kurt
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You have to make some assumption about how the coins is thrown. Let $O$ be the centre of the square. The simplest assumption is that the centre of the coin must land inside the square $S$ with centre $O$ and side 8, and the chance of its centre ending up in an area $A$ inside $S$ is $|A|/|S|$.

We have $|S|=64$. The coin will touch iff is centre lies inside a circle $C$ with centre $O$ and radius 4. You are lucky, because $C\subseteq S$, so $A=C$.

So prob = $16\pi/64=\pi/4\approx 0.78$.

almagest
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