$300$ dice rolls, at most $42$ "$2$'s". If a fair die is rolled $300$ times, what is the probability of rolling at most $42$ "$2$'s"? 
I plug this into my calculator: binomcdf($300, .166667, 42$) and get $.121$ as the solution. This is not one of my possible choices. What am I doing wrong?
 A: Use the normal approximation to the binomial, which works well when $np$ and $nq$ are large.
The expected number of $2$s is $\dfrac 1 6\cdot 300 = 50$.
The variance of the number of $2$s is $\dfrac 1 6\left(1 - \dfrac 1 6\right)\cdot 300 = 41+\dfrac 2 3$, so the standard deviation is about $6.453972244\ldots{}$.
Being $42$ or less is the same as being strictly less than $43$, so we use $42.5$.
If $X\sim N(50, 41+\frac 2 3)$, then
$$
\Pr(X\le 42.5) = \Pr\left( \frac{X-50}{6.453972244} \le \frac{42.5-50}{6.453972244} \right) = \Pr(Z\le -1.161895) = 0.122639. 
$$
A: FWIW here is my guess.  I emphasize that it's only a guess.  First note that with $n=300$ and $m=42$ we have
$$P(X\le m)=0.1212\ ,\qquad P(X=m)=0.0297\ .$$
If you were going to ask this as a multiple choice where the first expression above is correct, the second is an obvious distractor.  Other obvious distractors are $1$ minus the above, that is,
$$0.8788\ ,\qquad 0.9703\ .$$
Now these four numbers are not too far from your choices.  So my guess would be that $n$ and $m$ were originally different numbers, someone changed the question and forgot to change the answer.
A: using normal approximation to the binomial
np = 300 x 1/6 =50
standard deviation = sqrt( np(1-p)) = sqrt(50x5/6) = 6.455
using excel function
p(x<= 42) ==NORM.DIST(42,50,6.455, TRUE) = 0.10760709
