What property does this binomial probability calculation use? I have an equation
$$\sum_{k=2}^7{7\choose k}{0.01^k}(1-0.01)^{7-k} = 1-(0.99)^7 - 7(0.01)(0.99)^6 \approx 0.002031$$
I don't know what property the teacher used to quickly transform the summation to two simple equations. Can someone please give me a hint?
P.S. This formula is used to calculate binomial probability.
 A: I'm going to call the random variable of interest $X$ and that it follows a $\text{Bin}(7,.01)$.
Then the probability of interest appears to be
$$P(X\geq 2) = \sum_{k=2}^7\binom{7}{k}(.01)^k(.99)^{7-k}.$$
However, using complements results in an easier calculation:
$$P(X\geq 2) = 1-(X\leq 1)=1- \sum_{k=0}^1\binom{7}{k}(.01)^k(.99)^{7-k},$$
which is what you have.
A: The property is the Law of Total Probability, but you can also do it this way.
$$\begin{align}\sum_{k=2}^7 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}
 & = \Big(\sum_{k=0}^7 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big) - \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big)
\\[1ex] & = (p+(1-p))^7- \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big)
& \textsf{Binomial Expansion}
\\[1ex] & = 1- \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big)
\\[1ex] & = 1- {^7\mathsf C_0}\,p^0\,(1-p)^{7-0}-{^7\mathsf C_1}\,p^1\,(1-p)^{7-1}
\\[1ex] & = 1- (1-p)^{7}-7\,p\,(1-p)^{6}
\\[2ex]\sum_{k=2}^7 {^7\mathsf C_k}\,(0.01)^k\,(0.99)^{7-k}
& = 1 - (0.99)^{7}-7(0.01)(0.99)^{6}
\\[1ex] & = 0.002\,031\,041\,634\,94
\end{align}$$
A: $$
\sum_{k=0}^n \binom{n}{k} a^k b^{n-k} = (a+b)^n
$$
In your case, $a = 0.01, b= 1-a$ so $a+b = 1$. The sum is missing 2 terms ($k=0,k=1$) so the result is the full sum less those two terms
