Probability of a coin falling within a tile Consider the following question:
Question

I considered leaving a border of 3 cm on each side of the tile.
Image

Now if the center of the coin falls anywhere within this region, the coin stays inside the tile. Otherwise it moves, fully or partly, outside the tile.

So favourable area = 4*4 = 16

Area of sample space = 10*10 = 100

Probability = 16/100 = 0.16

But the correct answer is 0.36

Could anyone please explain?
(My previous answer was wrong - we all make mistakes, as suggested in this new response)

You seem to be correct to be concerned

Your calculation is effectively $\dfrac{(10-2\times 3)^2}{10^2}=\dfrac{4^2}{10^2}=0.16$ and looks sensible

There are two obvious ways to get $0.36 =\dfrac{6^2}{10^2}$, either as $\dfrac{(10-2\times 2)^2}{10^2}$  or as  $\dfrac{(2\times 3)^2}{10^2}$, where the former uses the wrong radius of the coin and the latter fails to subtract from the side of the square.  Both would be wrong, and I would guess that both errors could be possible
(My previous answer was wrong - we all make mistakes, as suggested in this new response)

You seem to be correct to be concerned

Your calculation is effectively $\dfrac{(10-2\times 3)^2}{10^2}=\dfrac{4^2}{10^2}=0.16$ and looks sensible

There are two obvious ways to get $0.36 =\dfrac{6^2}{10^2}$, either as $\dfrac{(10-2\times 2)^2}{10^2}$  or as  $\dfrac{(2\times 3)^2}{10^2}$, where the former uses the wrong radius of the coin and the latter fails to subtract from the side of the square.  Both would be wrong, and I would guess that both errors could be possible