How can I calculate $\mathbb{P}\left(\sum_{i=1}^5\left[\frac{Y_i}{3}\right]=3\right)$? Suppose $Y_1,Y_2,\ldots,Y_5$ is a random sample of distribution with Probability function $$\mathbb{P}(Y=y)=\frac{1}{3} , \quad y=1,3,5.$$
How can I calculate $\mathbb{P}\left(\sum\limits_{i=1}^5\left[\dfrac{Y_i}{3}\right]=3\right)$? 
Here $[y]$ denotes the greatest integer less than or equal to $y$.
 A: Since you tagged your question as self-study, I'll not give the full answer here but try to help you get on the right path.
To clarify, you're using $[\bullet]$ to denote the floor function. In which case, since $Y_i$ can take on values 1, 3, and 5, then $\left[ \frac{Y_i}{3} \right]$ can only be 0 or 1 (since 3/3 = 1 and 5/3 = 1.66... which would also get rounded down to 1). 
Think about what this means. You can now consider $\left[ \frac{Y_i}{3} \right]$ as your random variable, which can take on 2 values, with probabilities that you can work out from the original probability distribution for $Y_i$. 
Note that this essentially reduces the problem to a series of coin tosses with an unfair coin. 
A: The floor of $Y_i/3$ is 1 when $Y = 3,5$ and zero otherwise. So, for that sum to equal 3, you need exactly 3/5 of the $Y_i \in \{3,5\}$. So, the sum you wrote has a binomial distribution with $5$ trials and success probability $2/3$, so the answer is 
$$ \frac{5! \times 2}{3^6} $$ 
You can also just count the possibilities, as you will find there are 243 possible combinations and 80 of them produce the sum you said. 
