Find the probability that a stick will lie entirely on the tile. A floor is paved with tiles, each tile being a parallelogram such that the distance between pairs of opposite sides are $a$ and $b$ respectively, the length of diagonal being $L$. A stick of length $C$ falls on the floor parallel to the diagonal. Show that the probability that it will lie entirely on one tile is $$\left(1-\frac CL\right) ^2.$$
If a circle of diameter $d$ is thrown on the floor, show that the probability that it will lie on one tile is : $$\left(1-\frac da\right) \left(1-\frac db\right).$$
I tried this one hard. But unable to get the correct answer. 
What i did is - 
1. Since the stick is to be parallel to the diagonal , it can lie within the parallelogram and shapes like hexagonal. 
2. Since the altitude of the parallelogram is a and b. I assumed sides as x and y respectively. So, using area of parallelogram.
3. I ended with x = bk and y = ak. So the area of parallelogram comes out to be "abk".
4. But I am unable to find the ratio k. 


My drawing skills are not so good but I think this should help you give a visual intuition for the problem. 
If you have a stick of length $C$, parallel to the diagonal, you can see that the possible space where this stick lands is the trapezoid with one base the diagonal $L$ and the other base the stick of length $C$ touching both segments $a$ and $b$. 
If you write down the proportions: 
$$ \frac{AE}{AB} = \frac{DF}{DB} = \frac{C}{L} $$
From here you can see that: 
$$ EB = a - a\frac{C}{L}, BF = b - b\frac{C}{L} $$
And now in order to find the probabilities, you just have to find the proportion between the two areas: 
$$\frac{a(1-\frac{C}{L})b(1-\frac{C}{L})}{ab} = \big(1-\frac{C}{L}\big)^2$$
