The manager chooses 2 guests at random. Find the probability that they are both staying in the same room. There are $25$ rooms in a hotel; $3$ with $0$ guests, $4$ with $1$ guest each, $12$ with $2$ guests each, $5$ with $3$ guests each and $1$ with $4$ guests.
Total rooms: $25$
Total guests: $47$
Q. The manager chooses $2$ guests at random. Find the probability that they are both staying in the same room.
I got 
$$12\left(\frac{2}{47} \cdot \frac{1}{46}\right) + 5\left(\frac{3}{47} \cdot \frac{2}{46}\right) + 1\left(\frac{4}{47} \cdot \frac{3}{46}\right) = \frac{33}{1081}$$ 
Somebody explain this to me.

There are 
$$\binom{47}{2}$$
ways for the manager to select two of the forty-seven guests in the hotel.
In order to select two guests from the same room, the manager must select both guests from a room with two people or two of the three guests in a room with three people or two of the four guests in a room with four people.  Since there are $12$ rooms with two people, $5$ rooms with three people, and one room with four people, the manager can do this in 
$$12\binom{2}{2} + 5\binom{3}{2} + \binom{4}{2}$$
ways.  
Hence, the desired probability is 
$$\frac{12\dbinom{2}{2} + 5\dbinom{3}{2} + \dbinom{4}{2}}{\dbinom{47}{2}}$$
Note:  I assume you meant to write 
$$12\left(\frac{2}{47} \cdot \frac{1}{\color{red}{46}}\right) + 5\left(\frac{3}{47} \cdot \frac{2}{46}\right) + 1\left(\frac{4}{47} \cdot \frac{3}{46}\right)$$
in which case you would have obtained the correct probability.
