Probability answer cross check A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that
40% of this population prefers the color red. If
16 buyers are randomly selected, what is the probability that at least
2 buyers would prefer red? Round your answer to four decimal places.
My method:
o.40 * 16 / 2 = 3.2
Did I get it right? if not, what is the best approach
 A: Your calculation cannot be right. For any probability must be between $0$ and $1$, but $3.2$ is not.
We have repeated an experiment independently $16$ times. Each time the probability of success (prefers red) is $0.4$. We want the probability that the number of successes is $\ge 2$.
The probability that the number of successes is $k$ is $\binom{16}{k}(0.4)^k(0.6)^{16-k}$. For the probability that the number of successes is $\ge 2$, add up, $k=2$ to $16$. This will give you a correct expression, and with a great deal of work you can evaluate the $15$ terms and find their sum.
However, there is a simpler way. Let $p$ be the probability of $0$ or $1$ successes. Then the probability of $\ge 2$ successes is $1-p$. If we can find $p$, we can then easily compute the answer.
The probability of $0$ successes (nobody in our sample prefers red) is $(0.6)^{16}$. 
The probability exactly one person prefers red is $\binom{16}{1}(0.4)(0.6)^{15}$.
So $p=(0.6)^{16}+\binom{16}{1}(0.4)(0.6)^{15}$. Calculate, and to finish find $1-p$.
A: When you select $k$ persons from 16 buyers, the probability that exactly $k$ buyers would prefer red is $\binom{16}{k} 0.4^k (1-0.4)^{16-k}$.  Therefore,
\begin{align}
&P(\text{at least 2 buyers would prefer red}) \\
=& 1 - P(\text{0 buyer would prefer red}) - P(\text{1 buyer would prefer red}) \\
=& 1-\binom{16}{0} 0.4^0 (1-0.4)^{16-0} - \binom{16}{1} 0.4^1 (1-0.4)^{16-1} \\
=& 0.9967 \text{(cor. to 4 d.p.)}
\end{align}
