Show that the probability that it falls entirely on one brick is $\frac{(a-c)(b-c)}{ab},( c<a, c<b).$

Question

Question: A floor is paved with rectangular bricks each of length $a$ units and breadth $b$ units. A circular disc of diameter c is thrown on the floor. Show that the probability that it falls entirely on one brick is $\frac{(a-c)(b-c)}{ab},( c<a, c<b)$

I tried to solve but ended up getting $\frac{\pi^2}{4ab}.$ What am I missing?Help!

Answer 1

The area where the disk center can fall divided by the total area is:

$$\frac{(a-c)(b-c)}{ab}$$