A disk that is 1 cm in diameter is tossed onto a 6x6 grid of squares with each square having side length 1.25 cm.
A disk is in a winning position if no part of it touches or crosses a grid line- otherwise it is in a losing position
The disk is tossed and it lands in a random position so that no part of it is outside the grid.
What is the probability that the disk is in a winning position?
The radius is only 0.5 cm, so you win if the center of the disk is in an area of (6*.25)^2 = 2.25.
But the answer depends on what it means to land "onto" the 6x6 grid. Three interpretations occur to me.
1) The entire disk must be in the grid.
Then the total area of possible center locations = (7.5-1)^2 = 42.25
Probability of a win = 2.25/42.25 = 9/169 = approx 5.3%
2) The disk's center must be in the grid.
Then the total area of possible center locations = 7.5^2 = 56.25.
Probability of a win = 2.25/56.25 = 9/225 = 1/25 = 4%.
3) Some part of the disk must be in the grid.
Then the total area of possible center locations = (7.5+1)^2 = 72.25.
Probability of a win = 2.25/72.25 = 9/289 = approx 3.1%.
Fixed original entry to subtract radius of .5 from both sides of 1.25 square
Edited on July 8, 2016, 9:07 pm
Three answers (spoiler)
