Understanding the dependency of trials. Problem: A certain population is made up of 80% Mexican Americans.Select a jury of 12.Find the mean and standard deviation.
Problem Link: https://youtu.be/4Ew60JEPGUk?list=PL5102DFDC6790F3D0&t=865
The instructor assumes success as selecting a Mexican American.So the problem boils down to calculating the expected no of successes i.e if X is a random variable representing the number of successes then we have to find 
$E(X)$.
According to my understanding, we can use binomial distribution only if the trials are independent but in this problem selecting a person from the population changes one of the amounts and hence the probability of success so how is it feasible to apply binomial distribution here? Let me know if I'm not clear.
 A: The count for successes when selecting without replacement would be a hypergeometric random variable rather than binomial, though for selection of a small enough sample from a large population this approximates selection with replacement (if removing small amount of successes does not significantly impact the proportion in the population). 
However, this is of no import for the expectation due to the Linearity of Expectation
It seems counterintuitive the first time students encounter this, but expectation of a series of trials equals the series of expectations of each trial,  without any influence at all by whether the trials are independent or not.
If $X$ equals the count of successes in $n$ selections of jurors, and $X_k$ indicates a success on trial $k$-th selection, so $X=\sum_{k=1}^n X_k$, then the expectation for $X$ will be : $\mathsf E(X) ~{ =  \mathsf E(\sum_{k=1}^n X_k) \\ = \sum_{k=1}^n \mathsf E(X_k) \\ = n p}$ 
Reguardless of how you model the selection.

To get a idea of what is going on, consider the count in a sample of size two; whether selecting with or  without replacement, the expectation is:
$$\mathsf E(X) ~{= \mathsf E(X_1+X_2) \\ = \sum_{(x_1,x_2)\in \{0,1\}^2} (x_1+x_2)\mathsf P(X_1=x_1, X_2=x_2) \\ = \sum_{x_1=0}^1\sum_{x_2=0}^1 x_1\mathsf P(X_1=x_1, X_2=x_2) + \sum_{x_1=0}^1\sum_{x_2=0}^1 x_2\mathsf P(X_1=x_1, X_2=x_2) \\ \\ = \sum_{x_1=0}^1 x_1\sum_{x_2=0}^1 \mathsf P(X_1=x_1, X_2=x_2) + \sum_{x_2=0}^1 x_2\sum_{x_1=0}^1 \mathsf P(X_1=x_1, X_2=x_2) \\ = \sum_{x_1=0}^1 x_1\mathsf P(X_1=x_1)+ \sum_{x_2=0}^1 x_2\mathsf P(X_2=x_2) \\ = \mathsf E(X_1)+\mathsf E(X_2) \\ = 2p }$$
And so on for larger sample sizes.   (Use a proof by induction if you wish.)
A: Extended Comment: Comparing Hypergeometric and Binomial Distributions
Here are plots comparing hypergeometric and binomial distributions in two situations. 
Small population. The upper panel shows
a hypergeometric random variable (blue bars) counting the number of
Mexican-Americans chosen when $n = 12$ people are chosen
at random without replacement from a population of size $N = 100.$ of which 80 are
Mexican-Americans. The red bars are for $\mathsf{Binom}(12, .8),$ which
models sampling with replacement. 
The distinction between sampling without replacement (hypergeometric) and
with replacement (binomial) may be of practical importance.
[The hypergeometric distribution has slightly smaller variance.]
Large population. By contrast the lower panel shows results when 12 people are chosen
at random without replacement from a population of size $N = 1000,$ of which 800
are Mexican-Americans. Again here, the red bars are for $\mathsf{Binom}(12, .8).$

At the resolution of the graph, the distributions are not distinguishable
for the larger population.

In case it is of interest, here is the R code that produced the plots:
par(mfrow=c(2,1))

 # population of 100
 x = 0:12;  pdf = dhyper(x, 80, 20, 12)
 plot(x-.05, pdf, lwd = 2, type="h", ylim=c(0,max(pdf)), col="blue", 
   xlab="x", main="Pop of 100; Hypergeom PDF (blue) vs Binom (red)")
 abline(h=0, col="green2")
 pdf = dbinom(x, 12, .8)
 lines(x+.05, pdf, lwd=2, type="h", col="red")

 # population of 1000
 x = 0:12;  pdf = dhyper(x, 800, 200, 12)
 plot(x-.05, pdf, lwd = 2, type="h", ylim=c(0,max(pdf)), col="blue",
    xlab="x", main="Pop of 1000; Hypergeom PDF (blue) vs Binom (red)")
 abline(h=0, col="green2")
 pdf = dbinom(x, 12, .8)
 lines(x+.05, pdf, lwd=2, type="h", col="red")

par(mfrow=c(1,1))

