Probability of network computer If a computer blows up after turning it on with probability $0.05$. We have an order to create a network of $45$ computers, and we gathered $50$ computers for this purpose. What's the probability, that we can create a network with these computers?
We can calculate it by the simple way:
$\binom{50}{45}\cdot \left(0.95\right)^{45}\cdot \left(0.05\right)^2+...+0.95^{50}$
But my question is how we can calculate the probability by these 2 ways:

*

*Central limit theorem and Normal distribution.

*Approximation of the binomial distribution by Poisson distribution.

 A: Binomial Distribution
By the way, you computed the probability that you get 45 non-blown-up computers, but you can also create a network with 46 or more computers, so the actual exact answer is:
$\sum_{i=45}^{50}{50\choose i}(.95)^i(.05)^{50-i}$. This can be computed in R using the command
pbinom(44,50,.95,lower.tail=FALSE)

to get $0.9622238$.
Normal Approximation
A binomial distribution can be approximated by a normal distribution like the following: $Binomial(n, p)\approx Normal(np, np(1-p))$. So $Bin(50,.95)\approx N(47.5,2.375)$. You need to calculate the area to the right of a standard normal distribution of the z-value $\frac{45-47.5}{2.375}\approx -1.05$.
You can do either of the following commands, one being the standard normal distribution and the other specifying the mean and standard deviation:
pnorm(-1.62, lower.tail=FALSE)
pnorm(45,47.5,sqrt(2.375),lower.tail=FALSE)

to get $0.9473839$ and $0.9476213$.
Poisson Approximation
The poisson distribution can be a good approximation to the binomial distribution when n>20 and np<5, which both hold in our case. Then, $Bin(n,p)\approx Poisson(np)$. In this case, $Bin(50,.05)\approx Poisson(2.5)$. Now we are counting the number of defective computers, so we want this to run from 0 to 5. The answer is $e^{-2.5}\sum_{k=0}^5\frac{2.5^k}{k!}$
This can be calculated in R using the command
ppois(5,2.5,lower.tail=TRUE)

to get $0.957979$.
