Confused on Expected Value I have this question in which I have solved so many times but unsure of the answer i have. could anyone help out?

** A cpu manufacturer has a machine which produces microchips. The
  probability that the machine produces a faulted microchip is given by
  p = 1/10000. The manufactorer produces n = 6000 chips using the
  machine. Calculate the probabilty that there are 5 faulted chips, 6
  faulted chips, 20 faulted chips out of the n = 6000. Moreover, derive
  the expected number of faulted chips, and the square of the standard
  deviation.**

My solution gave me 461/10000 as the expected value. Is that correct?
Regards
 A: The number $X$ of faulty chips in a sample of $6000$ has Binomial Distribution, parameters $p=\frac{1}{10000}$, $n=6000$.
The mean of $X$ is $np$, and the variance (square of the standard deviation) is $np(1-p)$. 
So to answer your specific question, the mean of $X$ is $(6000)(1/10000)$. 
The probability that there are exactly $k$ faulty chips in the sample is 
$$\binom{6000}{k}p^k (1-p)^{6000-k}.$$
You can use this formula to compute the various probabilities asked for.
However, it is possible that you are expected to use the Poisson approximation to the Binomial. If the word Poisson is not familiar to you, you can stop reading. If it is familiar, continue.
In this sort of situation, where $p$ is small, $n$ is large, and $np$ "moderate," we can approximate the distribution of $X$ using a Poisson random variable with parameter $\lambda=np=0.6$.
The probability that $X=k$ is well approximated by $e^{-\lambda} \dfrac{\lambda^k}{k!}$. 
A: Give the chips an index and let $X_{i}$ take value $1$ if chip $i$
is faulted and $0$ otherwise. 
Then $$X=X_{1}+\cdots+X_{6000}$$ stands for the total number of faulted
chips.
We find: $$\mathbb{E}X=\mathbb{E}X_{1}+\cdots+\mathbb{E}X_{6000}=\frac{1}{10000}+\cdots+\frac{1}{10000}=\frac{6000}{10000}$$
