What property does this binomial probability calculation use?

Question

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.

Answer 1

The property is the Law of Total Probability, but you can also do it this way.

$$\begin{align}\sum_{k=2}^7 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k} & = \Big(\sum_{k=0}^7 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big) - \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big) \\[1ex] & = (p+(1-p))^7- \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big) & \textsf{Binomial Expansion} \\[1ex] & = 1- \Big(\sum_{k=0}^1 {^7\mathsf C_k}\,p^k\,(1-p)^{7-k}\Big) \\[1ex] & = 1- {^7\mathsf C_0}\,p^0\,(1-p)^{7-0}-{^7\mathsf C_1}\,p^1\,(1-p)^{7-1} \\[1ex] & = 1- (1-p)^{7}-7\,p\,(1-p)^{6} \\[2ex]\sum_{k=2}^7 {^7\mathsf C_k}\,(0.01)^k\,(0.99)^{7-k} & = 1 - (0.99)^{7}-7(0.01)(0.99)^{6} \\[1ex] & = 0.002\,031\,041\,634\,94 \end{align}$$

Answer 2

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.

Answer 3

$$ \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} = (a+b)^n $$

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