How to find $P[R\ge 13]$?

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R follows a binomial distribution such that $R$~$Bin(24,0.4)$

I don't understand how to get the answer for this. From my understanding, $$P[R\ge 13]=P[R=0]+P[R=1]+P[R=2]+P[R=3]+..P[R=13] $$, doing this I don't get 0.003291, I got 0.1142651=not the solution.

The correct answer is 0.003291 but I keep getting the wrong answer. Is the way I did it wrong?

Edit: Here is the whole question Red and white chocolate balls are randomly filled into boxes containing a total of 24 balls. The color of each ball is determined by a random mechanism such that on average 60% of all balls are black.

(a) Let the random variable R denote the number of red chocolate balls in a box. What is the distribution of R?

R is binomial with parameters n= 24, p=0.4

(b) What is the probability that all chocolate balls in one box have the same color?

P[R = 24] + P[R = 0]= 4.74x$10^{-10}$

(c) How many boxes need to be selected to have a chance of > 90% to have at least one box with at least 13 red chocolate balls?

So I know how to find the number of boxes but it seems I made a mistake at one point, which led me to the wrong answer.

$P[R\ge 13]= 0.003291$ is the correct answer for part one

Why? I don't understand how they got that answer.

Then I need to $$P\left[X\:\ge \:1\right]\:=\:1−P\left[X\:=\:0\right]\:=\:1−\begin{pmatrix}m\\ 0\:\end{pmatrix}\left(p'\right)^0\left(1−p'\right)^m\:=$$

$$\:1−0.9967^m>0.90$$

Then I need to use logs to find m.

But I went wrong at $P[R\ge 13]$ and I don't understand how to find it.

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(a) R has binomial distribution $B(n=24,p=0.4)$

(b) $P(R=0)+P(R=24)=\binom{24}{0}\times0.4^0\times0.6^{24}+\binom{24}{24}\times0.4^{24}\times0.6^0\approx4.74\times10^{-6}$

(c) Chance for a box to have at least 13 red chocolate balls $$P(R\geq13)=\sum_{k=13}^{24}{P(R=k)}\approx0.1143$$ then we answer the question 'after how many $p=0.1143$ trials the chance of only failures is less than 10%', also using a binominal distribution.

$pmf_B(k=0,n=x,p=0.1143)<0.1$

$\binom{x}{0}\times0.1143^0\times(1-0.1143)^{x-0}<0.1$

$0.8857^x<0.1$

$x>\log_{0.1}{0.8857}$

The answer is 19.