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, $$P[R\ge 15]=P[R=0]+P[R=1]+P[R=2]+P[R=3]+..P[R=15] $$, 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.
$P[R\ge 15]= 0.003291$ is the correct answer. Why? I don't understand how they got that answer.
I find $P(R=0)=0.6^{24}\approx 4.74\cdot 10^{-6}$, much larger than your calculation. Your equation for $P(R \ge 15)$ has all the wrong terms on the right. The right side computes $P(R \le 15)$
I find the probability of at least $13$ red balls to be about $0.1143$, which matches your number that you say is $P(R \ge 15)$, but I am computing $P(R \ge 13)$. If you open $n$ boxes, the chance none of them have at least $13$ red balls is then $(1-0.1143)^n$, so to have $90\%$ chance of at least one having $13$ red balls you solve $(1-0.1143)^n=0.1$ and find $n$ is $19$.
I don't find any $0.003291$ in the binomial distribution here, either point values or cumulative.