Probability Question with random sample i am trying to solve some probability question. Here are they


*poll found that 73% of households own digital cameras. A random sample of 9 households is selected
A. What is the probability that 2 of the households own a digital
camera?
B.What is the probability that at most 1 of households owns a digital camera?
C. What is the probability that at least 1 household owns a digital camera?
D. Now suppose a random sample of 450 households is selected. Use R to find the probability that at least 350 own a digital camera
I have answered A, B and C using R
P0<-(factorial(9))/((factorial(0)*factorial(9-0))) * (0.73^0) * ((1-0.73)^(9-0))
P1<-(factorial(9))/((factorial(1)*factorial(9-1))) * (0.73^1) * ((1-0.73)^(9-1))
P2<-(factorial(9))/((factorial(2)*factorial(9-2))) * (0.73^2) * ((1-0.73)^(9-2)) 
list("Number11A" = P2, "Number11B" = P0+P1, "Number11C" = 1-P0,"Number11D Mean" = 9*0.73,"Number11D Standard Deviation"= sqrt(9*0.73*(1-0.73))) #Number_11A

I found that A , B and C will be ( 0.002006756,0.0001931818 , 0.9999924)
However, I cannot understand question D as this question has a sample . Thank you very much for your help. 
 A: You've correctly combined the relevant pprobabilities based on $\binom{9}{0}$, $\binom{9}{1}$, and $\binom{9}{2}$ for parts A through C.
Now, as I think you've realized, you need to find something using $\binom{450}{350}, \binom{450}{351}, \dots, \binom{450}{450}$.
Using the formula $\binom{450}{350} = \frac{450!}{350! 100!}$ is not great, because, well, $450!$ is big. Really big. Like $10^{1000}$ big. That's a googol to the 10th power. The standard formats computers use for representing numbers can only reliably go up to about $10^{176}$; $450!$ is way bigger than that.
So you'll need to do it a different way.
A few possibilities:


*

*Figure out how to compute $\binom{450}{n}$ without directly using $450!$.

*Realize that this is probably a common problem, and R probably already has a way built in to compute large binomial coefficients.

*There's actually a more direct function in R to do exactly what you're looking for. What would it be called?

A: In problems posed like that one, without the size of the population from which the sample was taken, one must assume each of the responses $X_i$ as an independent realization of a Bernoulli with $p=.73$, so the number $X$ of households in the sample with a camera is the sum $X=X_1+X_2+\ldots+X_{450}$ of the individual responses, which will have a Binomial distribution with $n=450$ and $p=.73$.
One way to solve this question without resorting to direct computation of factorials would be the normal approximation, which gives $$P(X\ge350)\approx P(Y>350-.5),$$ where $Y$ has Normal distribution with $\mu=np=450\times.73=328.5$ and $\sigma^2=np(1-p)= 450\times.73\times.27=88.695,$ and where the "$-.5$" is due to Yates' continuity correction.
So, $$P(X\ge350) \approx P(Y> 350-.5) = P(Z>(349.5-328.5)/\sqrt{88.695}) \approx P(Z>2.23),$$ which can be obtained from linear interpolation from Standard Normal distribution table.
