# Probability question about whether the landing spot of a tossed coin covers a specified point in space

I am having some trouble with the following question from Chapter 5 review of Pitman's book titled Probability:

1. A coin of diameter 1 inch is tossed in the air and caught in an empty soup can of bottom radius 3 inches. The coin lies flat on the bottom.

a) What is the chance that the coin covers the center point of the bottom of the can?

Suppose that instead of the soup can, the coin is dropped into a box whose bottom is a square with sides of length 5 inches.

b) What is the chance that the coin covers the center point of the bottom of the box?

c) Consider one of the main diagonals of the bottom of the box. What is the probability that part of the coin crosses that diagonal line?

I was able to arrive at the correct answer for part (a) by using the following logic: Consider the center of the coin. The center of the coin must land inside the circle of radius 2.5 inches inside the larger circle of 3 inches that is the bottom of the soup can (with the same center). The coin covers the center of the can's bottom iff the landing spot of the coin's center falls within the circle of radius 1/2 at the center of the can. Assuming the landing spot of the coin's center is distributed uniformly on the circle of radius 2.5 inches, the answer to (a) is simply (area of circle of radius 1/2)/(area of circle of radius 5/2) = 1/25

I tried to use the same logic was (b) and (c) but was unable to arrive at the same answer. Namely, I assumed the probable landing spot of the center of the coin is distributed uniformly on the square of side 4 inches and divided the area where the coin's center can land in order for the event to happen, by the area of the square.

FYI, the answers to (b) and (c) are 0.039 and 0.29 respectively.

I suspect where my logic is flawed is the assumption that the density of the coin's center's landing spot is distributed uniformly over the smaller square, since there is a small area on each of the 4 corners of the larger square that cannot possibly be covered the coin. But I am still stumped at the moment.

• I'm wondering if the two answers are incorrect: I'd previously posted a comment that matched (b) result (same technique matched (c) result), then realized I'd made an error that affected both. Correcting that gives me, e.g., Pi/64 for (b), same result directly doing the calcs against the distributions in Mathematica, and same result simulating it. What results are you getting? Jan 20, 2015 at 20:37

It appears the answer given to you (or in the book solutions?) is incorrect. The correct answer, e.g., for (b) is $\frac{\pi }{64}$ ~ $0.049$, and the correct answer for (c) is $\frac{1}{32} \left(8 \sqrt{2}-1\right)$ ~ $0.32$

I'd venture the author made the same boo-boo on both that I had in my earlier (now deleted) comment.

• Thanks, those were indeed the answers I was getting! The ones I included in the question were the book's solutions. At least now I know somebody else is getting the same thing I did! Jan 22, 2015 at 5:17
• @Economist: Glad to have helped - I know the frustration of burning time on wrong answers / typos in texts! If the answer is satisfactory, please consider accepting/voting so future readers don't wonder if it's correct. Jan 22, 2015 at 5:55
• Oh right, thanks for reminding! First time using Stack Exchange... Jan 23, 2015 at 5:58
• Yes, I came up with the same conclusion, particularly because the problem in question is sandwiched between four others (two on each side) that are explicitly about uniform distributions. But after I saw that the official answer (in the hardcover edition) was what the OP wrote... I chickened out and deleted my comment. After a bit of googling today, this was however part of the known errata for the book: slc.berkeley.edu/sites/default/files/docs/… which has rather numerous errors for such a popular textbook... Jan 23, 2015 at 9:17