probability - find the variance of an event X Roll 8 fair dice. $X$ is the number of dice that land on $6$. How would you calculate the variance of $X$? 
Can we treat this problem as a binomial such that 
$p$ = probability of landing on a $6$, $p = \frac{1}{6}$ and 
$q$ = probability of not landing on a $6$, $q=\frac{5}{6}$. 
Then the variance  $\operatorname{Var}(X) = n p q = 8 \times \frac{1}{6} \times \frac{5}{6}$. 
Would this be correct?
 A: As I said in the comment, your solution is correct. Here is a simulation confirmation using Mathematica. The function RandomChoice is used to generate a million repetitions of 8 rolls. Command Tally shows the number of different outcomes generated:
In[56]:= sample = 
  Map[Count[#, 6] &, RandomChoice[Range[6], {10^6, 8}]];

In[57]:= Tally[sample]

Out[57]= {{0, 232113}, {3, 104193}, {2, 260399}, {1, 372385}, {4, 
  26287}, {5, 4152}, {6, 441}, {7, 30}}

In[58]:= {Variance[sample], 8 1/6  5/6} // N

Out[58]= {1.1125, 1.11111}

A: The calculation is absolutely correct.  Suppose that you repeat an experiment independently $n$ times, and the probability of "success" each time is $p$. If the random variable $X$ is the number of successes, then $X$ has binomial distribution, with mean $np$ and variance $npq=np(1-p)$.
The above dice problem fits the requirements perfectly.
There is a problem with the title of the post. You are finding the variance of a random variable, not of an event. The concept of variance does not even apply apply to events. An event is just a subset of the sample space.  So getting an odd number of $6$'s, for example, is an event. 
