Probability distribution (Binomial distribution) Let's say we have a coffee shop that only sell coffee and tea. 
Based on the past selling records, 40% of the customers order tea.(customer will either order tea or coffee - not both)
Now, for the simplicity, assume a random sample of 10 customers.
According to my understanding, the probability mass function will be
$P(X=k) = \pmatrix{10\\k} (0.4)^k (1-0.4)^{10-k}$ 
(taking that the $X$ is the number of customers order tea)
$X\sim B(10,0.4)$
In this case, if I want to find the probability that there are at most 2 customers ordering tea, I will just have to substitute $k=0, k=1, k=2$ into probability mass function and add them together.
However, what if I am told that there are at most 6 customers (in the same sample of 10) that order tea, and I want to find out the same probability that I find above ( probability of at most 2 customers ordering tea), will it be the same or the condition given above will affect on my probability? 
If so, how? If not, will there be any conditions that the given pre-condition(there are at most 6 customers that order tea) will affect the result of my probability?
 A: Let's define $E_i = \{ i$ of the 10 customers order tea $\}$.  
When you're told that at most 6 customers order tea, the probability of the event 
$$\{ \textrm{ at most 2 customers order tea }\} = \cup_{i=0}^2 E_i $$
 will change.  This is because you have removed the outcomes $E_7, E_8, E_9$, and $E_{10}$ from the sample space.  These are all possibilities that fall outside your event $\cup_{i=0}^2 E_i$, so the relative weight of your event increases; your probability should rise.
Two ways to approach this are (i) calculate the conditional probability $$ P(A | B) = {P(A \cap B) \over P(B) } $$ where $A = \cup^2_{i=0} E_i$ and $B = \cup_{i=0}^6E_i$.
(ii) Recognize that 4 customers are certain to buy coffee, so their purchases can be disregarded.  Now you consider only the six customers who have a choice to buy tea.  What is the probability that at most 2 of 6 customers orders tea?
A: By definition of conditional probability $$P(X \leq 2 | X \leq 6) = \frac{P(X \leq 2 \cap X \leq 6)}{P(X \leq 6)}$$
But $P(X \leq 2 \cap X \leq 6) = P(X \leq 2)$ since the set of events such that $X \leq 2$ is a subset of the set of events such that $X \leq 6$.
Therefore $$P(X \leq 2 | X \leq 6) = \frac{P(X \leq 2)}{P(X \leq 6)}$$
Overall this has the effect of increasing the probability that at most 2 customers order tea (compared to the unconditional case) since there are now only 6 customers rather than 10 who could potentially order tea. Intuitively, dividing by $P(X \leq 6) < 1$ scales this probability up by the appropriate factor.
A: From the fact that there are at most 6 customers (in the same sample of 10) that order tea, you are now sure that 4 out of 10 customers order coffee. Therefore, you will have a new random sample of 6 customers instead of 10.
The new probability mass function is then
$$P(X' = k) = \pmatrix{6\\k} (0.4)^k (1-0.4)^{6-k}.$$
A: You have correctly identified this as a binomial distribution.
You have correctly calculated the unconditional probability $\mathsf P(X\leq 2) =\sum_{k=0}^2 {10\choose k}0.4^k0.6^{10-k}$
For the conditional probability, just apply the definition.
$\begin{align}
\mathsf P(X\leq 2 \mid X\leq 6)
 & = \dfrac{ \mathsf P(X\leq 2\cap X\leq 6)}{\mathsf P(X\leq 6)}
\\[1ex]
 & = \dfrac{\mathsf P(X\leq 2)}{1- \mathsf P(X> 6)}
\\[1ex]
 & = \dfrac{\sum_{k=0}^2 {10\choose k} 0.4^k 0.6^{10-k}}{1-\sum_{k=7}^{10} {10\choose k} 0.4^k 0.6^{10-k}}
\\[2ex]
& \vdots
\end{align}$
Remark  In the second step, note that if a number is both at most 2 and at most 6, it is simply at most 2.
