Combinations and odds EDITED: I'm sorry about the confusion. I honestly forgot to include that information. So yes, the balls are chosen without replacement.
There are a group of 80 balls. All balls are colored. There are 5 different colors. Each colors appears in 16 balls. 2 balls are chosen randomly. What are the odds of getting both balls of the same color?
Total balls: 80
Total colors: 5
Total balls per color: 16 (5 x 16 = 80)
Chosen balls: 2

 A: This solution assumes that the balls are chosen without replacement. In other words, I'll assume that the second ball is chosen from a pool of $79$ possibilities once the first ball is removed. 

Suppose you've picked the first ball already. 
There are $79$ remaining balls. Of these, $15$ are the same color as the first, and the rest are not. This means that the probability of the balls being the same color is $$\boxed{\frac{15}{79}\,}$$ 
A: Suppose the colours are blue, red, yellow, green, and purple.
The probability of both balls being blue is that of the first ball being blue and the second ball being blue, which is $K=\frac{16}{80}\times\frac{15}{79}$.
The probability of both balls being red, or yellow, or green, or purple is also $K$.
The required probability is the sum of all these probabilities, which is $K+K+K+K+K=5K$.
A: You pick the first ball and note its colour.
Now we pick a second ball there are $80 - 1 = 79$ balls remaining of which $ 15 $ balls are the same colour.
So the probability of picking the same colour ball is
$$\dfrac{15}{79}$$ 
Now you didn't ask about probability but odds and these are not the same but related.  There are $ 79 - 15 = 64$ losing balls remaining and $ 15 $ winning ones so the odds are $ 64:15 $ against. 
