You've a coin of diameter 2cm, and you throw it on to the chessboard. The center of the coin falls somewhere on the chessboard.

What is the likelihood that the coin is completely within a white square?

Each white square of the chessboard of 80cm to a side would be (10cm)² in area. The area where a 2cm diameter coin would be within the white square would be (10cm - 2cm)² = 64cm². With 32 white squares on the chessboard, the total area within a white square for the coin to completely land is then just shy of 64cm²*32 = 2048cm².

The probability, then, depends on the size of the total area in which the 2cm diameter coin may land:

(Case 1)  If the coin is required to be totally within the boundary of the chessboard, then the total area would be just shy of
(80cm - 2cm)² = 6084cm².  The probability is then just shy of 2048/6084, i.e., approximately 33.662%.

(Case 2)  If the coin need only have its center fall within the boundary of the chessboard, then the total area would just shy of 80cm)² = 6400cm².  The probability is then just shy of 2048/6400, i.e., approximately 32%.

(Case 3) If the coin need only be touching a portion of the chessboard to be considered "falling somewhere on the chessboard", then the total area would be just shy of (80cm + 2cm)² = 6724cm².  The probability is then just shy of 2048/6724, i.e., approximately 30.458%.

As the problem gives that the center must fall within boundaries of the chess board, Case 2 is our answer.

(Case 4) From a point made by bob909 in a prior post, where the coin lands on edge. Assuming the thickness of the coin has no dimension and we ignore the probabilities on the orientation of how the coin may land, the probability is just shy of 3200/6400 = 50%.  Yet, if the diameter given is also the coin's thickness, we are back to the solution given in Case 2.

Edited on February 26, 2008, 6:14 am

Posted by Dej Mar on 2008-02-25 23:44:02