Why do we have to calculate 1-binomcdf for P(X>11)? Why do we have to calculate 1-binomcdf for P(X>11)?
Can't we just calculate binomcdf with left and right bound like the normalcdf?


 A: You have a random variable $X \sim \mathsf{Binom}(n = 12,\,p = 1/4)$ and you seek $P(X > 11).$ There are many ways to approach this
problem (with or without whatever software uses the function binomcdf).
I will show three main approaches. I think the first is best because it is so
simple. But I guess the author may have expected you to use the last.
(a) Using the binomial PDF formula:
$$P(X > 11) = P(X = 12) = {12 \choose 12}(1/4)^{12}(1 - 1/4)^0 = (1/4)^{12} = 
 5.9605 \times 10^{-08}.$$
Or in R statistical software, where dbinom is a binomial PDF:
dbinom(12, 12, .25)
## 5.960464e-08

(b) Using the binomial CDF: This would be tedious to compute by hand, but some
books have limited binomial CDF tables, and this choice of parameters may be included
there.
$$P(X > 11) = 1 - P(X \le 11) = 1 - [P(X=0) + P(X=1) + \cdots + P(X = 11)].$$ 
In R, where pbinom is a binomial CDF, the syntax is somewhat similar to what you showed in your question.
1 - pbinom(11, 12, 1/4)
## 5.960464e-08

(c) Using normal approximation: Some texts may ask you to evaluate this using a normal approximation to the binomial, but that approximation will work rather poorly for this problem because $n=12$ is relatively small and $p = 1/4$ is relatively far from $1/2.$ 
Here is how the normal approximation would be attempted. First, 
$\mu = E(X) = np = 12(1/4) = 3.$  Also,
$\sigma = SD(X) = \sqrt{np(1-p)} = \sqrt{9/4} = 3/2.$ Then
$$P(X > 11) = P(X > 11.5) = 
P\left(\frac{X - \mu}{\sigma} > \frac{11.5-3}{1.5}\right)
\approx P(Z > 5.67) \approx 0,$$
where $Z$ has the standard normal distribution, which places almost
no probabiltiy above 5.67. The normal approximation "works" only because
the result is so tiny that the mis-match is invisible. 
1 - pnorm(5.67)
## 7.139876e-09

In the figure below the exact binomial probabilities are shown as vertical blue
bars and the (ill-fitting) PDF of the normal approximation is shown as a black
curve. The probability of interest is to the right of the vertical broken red
line; this is a region where neither the binomial nor the normal distribution
has appreciable probability.

Note: I spent time with the normal approximation because the problem seems to have been contrived so that $\mu$ and $\sigma$ are very easy to calculate. This makes
me suspect that the normal approximation may be the intended method.
