Disk on a tile floor

Question

A disk 2 inches in diameter is thrown at random on a tiled floor, where each tile is a square with sides 4 inches in length. Let C be the event that the disk will land entirely on one tile. In order to assign a value to P(C), consider the center of the disk. In what region must the center lie to ensure that the disk lies entirely on one tile? If you draw a picture, it should be clear that the center must lie within a square having sides of length 2 and with its center coincident with the center of a tile. Since the area of this square is 4 and the area of a tile is 16, it makes sense to let $P(C) = \frac{4}{16}.$

I don't really understand the statement for the last sentence "Since the area of this square is 4 and the area of a tile is 16, it makes sense to let $P(C) = \frac{4}{16}.$"

Answer 1

The inner red square is the area of a tile in which the center of the disk could land and the disk would be contained entirely within the tile. The grey area is the area in which the center of the disk would land and the disk would not be entirely contained within the tile.

The area of a square is the side length, squared. The inner square has area $2^2 =4$, and the entire square (red and grey regions combined) has area $4^2 = 16$.

Therefore, the probability that the disk's center lands in the red zone is the area of the red zone, divided by the total area, which is

$$\frac{4}{16} = \frac{1}{4}$$

Answer 2

The square ABCD in the figure below represents one tile. The square EFGH is the region the center must lie in to keep the whole disk within ABCD. A circle centered at F is just tangent to two sides, showing that if the center were any further up or right the disk would extend outside the square. ABCD has side $4$ and area $16$, EFGH has side $2$ and area $4$.

Answer 3

The formula of probability of the event $C$ in this case will be:

$$P(C) = \frac{\text{Favourable area}}{\text{Total area where disc's centre can land}} = \frac{4}{16}$$