# 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 $$y$$.

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.

• The floor function is another name for the greatest integer less than or equal to y. Which is the way the OP stated it. – Michael R. Chernick May 2 '17 at 14:18
• Why then do you need to ask the OP to clarify that notation, if it is stated? – wolfies May 2 '17 at 16:45
• Fair enough. I guess it wasn't really intended as a question to the OP, more as a clarifying remark (just because the floor function is usually denoted $\left\lfloor\bullet\right\rfloor$). Edited my answer so it's no longer a question. – Ruben van Bergen May 3 '17 at 13:36

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.