Probability that fewer than 24 people logging into the site will make a purchase The problem is below: 

Suppose that the probability that one person logging into an e-commerce site will make a purchase is 0.1. If there are 200 people
  logging in the site in the next day, what is the probability that less
  than 24 people will make a purchase? Please select the correct
  answer:
A. 0.8264
B. 0.1735
C. 02461
D. 0.5

I tried using the Binomial Distribution formula in R to calculate the sum of probability that 1 - 23 people make a purchase: 
dist <- 0
for (i in seq(1,23)){dist <- dist + dbinom(i,200,0.1)}

And the result is 0.7982976, which is none of the answer choices. 
Does anyone know what is wrong with my solution? How should I solve this problem?
 A: I agree with @whuber that none of the answers is exactly correct. 
However, if (oblivious to a continuity correction) you take the key probability to be $P(X < 24),$ round excessively, and
do a normal approximation to binomial using printed tables, you get $0.8264.$
First, $\mu = np = 200(.1) = 20$ and $\sigma = \sqrt{np(1-p)} = 0.8250.$
Then
$$P(X < 24) = 
P\left(\frac{X-\mu}{\sigma} < \frac{24-20}{0.8250}\right)\\ \approx P(Z < 0.94) = 0.8264,$$
where $Z$ is standard normal.
You have to make just the right 'minor' errors in just
the right order to get this answer. Unfortunately, there are 
textbooks that show these errors in examples. And
there are multiple choice tests, in which one has
to get used to picking the closest answer, instead
of the exactly correct one. 
A: None of the answers is correct.
Evidently the answer ought to be substantially greater than $1/2=0.5$ because the mean of this Binomial distribution is $20=200\times 0.1$ and the event "less than 24" includes all values less than the mean and a substantial number greater than it.  This alone indicates (A) is the intended response.
The correct probability is given by the cumulative value for the Binomial$(200, 0.1)$  distribution evaluated at $23.$  In R you can find this directly or add the individual probabilities, thus:
> pbinom(23, 200, 0.1)
[1] 0.7982976

or
> sum(dbinom(0:23, 200, 0.1))
[1] 0.7982976

Often, textbook questions of this nature are expecting you to use a Normal approximation to the Binomial distribution.  This approximation is based on matching the Binomial mean, $\mu=200\times 0.1,$ and the Binomial variance, $\sigma^2 = 200\times 0.1 \times (1-0.1),$ to a Normal distribution and evaluating its cumulative value at $23 + 1/2.$  In R this would be
> pnorm(23 + 1/2, 200 * 0.1, sqrt(200 * 0.1 * (1-0.01)))
[1] 0.7842322

As you can see, this is a good approximation but it still does not agree with any of the answer choices.  I cannot find any variation of either approach that gives any of the answer choices.  The closest I have been able to get is to use the exact Binomial calculation and average its values for $23$ and $24:$
> mean(pbinom(23:24, 200, 0.1))
[1] 0.8267018

However, this still does not agree with any of the choices exactly.

As remarked in a comment, your method embodies a conceptual error: the event "less than 24" includes the possibility of zero, which you haven't included.  However, zero is so unlikely that omitting it produces no appreciable error; its chance is
> dbinom(0, 200, 0.1)
[1] 7.055079e-10

This error could matter in other problems, though, so it's worth remembering.
