Question: The proportion of the American adult population that supports candidate Green is p=0.22. A SRS of 9 adults asks if they agree with the statement “I support candidate Green.” What is the probability that at least 2 of those surveyed would agree with that statement? 

Ok, so far I know that the probability of at least 1 person would be .8931 -edited- was using .88 instead .78- Is that useful in solving for at least two?

Answer: 
The probability that at least $2$ support Green is $1$ minus the probability that $0$ people or $1$ person supports Green.

The probability that no one supports Green is $(0.78)^9$.

The probability that **exactly** $1$ person supports Green is $\dbinom{9}{1}(0.22)(0.78)^8$. 

We could also do the problem the hard way. The probability that exactly $k$ people support Green is $\dbinom{9}{k}(0.22)^k(0.78)^{9-k}$. Calculate all of these, for $k=2$ to $9$, and add up. 

**Remark:** We explain why the probability there is exactly one Green supporter is $9(0.22)(0.78)^8$. Line up the people we are interviewing. The probability that the *first* is a supporter of Green and the rest are not is $(0.22)(0.78)^8$. The probability that the first does not support Green, but the *second* does, and the rest don't is $(0.78)(0.22)(0.78)^7$, which is $(0.22)(0.78)^8$. Similarly, the probability the first two are not supporters, but the *third* is a supporter, and the rest are not is $(0.22)(0.78)^8$. We continue this way through all $9$ people, and end up with $(0.22)(0.78)^8$ added to itself $9$ times, for a total of $(9)(0.22)(0.78)^8$. The point is that there are $9$ different ways in which there is $1$ supporter of Green and $8$ non-supporters.
$$ \mathcal P( \text{at least two agree}) = 1 - \mathcal P(\text{ eight or nine disagree}) = 1- {9 \choose 8}p^8(1-p) - {9 \choose 9} p^9 = 1- p^8(9-8p),\quad p = .78$$