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. 
 A: Focus on one endpoint of the stick (say, the left endpoint). Since the orientation of the stick is fixed, everything you need to know about the position of the stick can be reduced to the location of that endpoint.
Think about where that endpoint can lie in the parallelogram, and which positions are "valid" (they allow the entire stick to lie in the parallelogram). You will find that the space of "valid" positions for that endpoint forms a similar parallelogram with diagonal $L-C$ (if the endpoint lies outside of this small parallelogram, the stick will not lie entirely in the parallelogram). Thus the probability is the ratio of the area of this smaller parallelogram to the area of the larger parallelogram: $\frac{(L-C)^2}{L^2} = \left(1 - \frac{C}{L}\right)^2$.

A similar technique can be used for the other question. Here, perhaps track the center of the circle. Where can the center of the circle lie if you want the entire circle to lie in the parallelogram? The center must be at least distance $d/2$ away from each edge, so the region of "valid" positions for the center is a similar parallelogram whose altitudes are $a-d$ and $b-d$.
A: 
The figure elaborates the solution given by angryavian above for the first question.
Here, I have considered the distribution of the mid-point of the stick so that it falls within the tile.
NOTE: GH, NB, JI and DM are the extreme positions of the stick within the tile with respective centres E, L, F and K.
Then the diagonal of the smaller parallelogram will have length $(L-C)$. Then the required probability will be the ratio of areas of the smaller parallelogram $ELFK$ to the larger parallelogram $ABCD$
As mentioned by angryavian, 
$Probability = \frac{area(ELFK)}{area(ABCD)}$
$= \frac{(L-C)^2}{L^2}$
$= (1-\frac{C}{L})^2$

The figure elaborates the solution given by angryavian above for the second question.
Here, the smaller rectangle $EFGH$ consists of all the points which could be the centre of the circle, so that the circle falls within the tile. 
From the above figure, it is clear that the required probability is Area(EFGH)/Area(ABCD)
$Probability = \frac{(a-d)(b-d)}{ab}$
$= (1-\frac{d}{a})(1-\frac{d}{b})$
A: 
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: 
$$\frac{a(1-\frac{C}{L})b(1-\frac{C}{L})}{ab} = \big(1-\frac{C}{L}\big)^2$$
