I have to calculate a binary distribution. Normally I would use a table in the back of my Statistics textbook or just crunch the numbers on a few (or quite a few) terms from the sum, but the latter is tedious. The table only has sums from $k=0,n$ where $0<p\le0.5$. I have $p=0.7$.
My question is, how can I rewrite the formula so that the $p$ I am working with is less than or equal to $0.5$? I know the following is true for my problem:
$$P(Y_7\leq \pi_{0.7})=\sum_7^\infty \binom{8}{k}(0.7)^k(0.3)^{8-k}=1 - \sum_0^6 \binom{8}{k}(0.7)^k (0.3)^{8-k}$$
This is NOT hard to solve manually, but I would appreciate, moving forward, a means to convert $p=0.7$ into $p=0.3$ so I can use the table instead of crunching all the terms myself. I have tried several times and I am missing something. Thank you.
You've mistaken the word "binary" for "binomial".
You have $$ \sum_{k=7}^8 \binom 8 k (0.7)^k (0.3)^{8-k} = 1 - \sum_{k=0}^6 \binom 8 k (0.7)^k (0.3)^{8-k} $$ but also $$ \sum_{k=7}^8 \binom 8 k (0.7)^k (1-0.7)^{8-k} = \sum_{\ell=0}^1 \binom 8 \ell (1-0.3)^{8-\ell} (0.3)^\ell. $$ If $Y \sim \operatorname{Binomial}(8,0.7)$ then $8-Y\sim\operatorname{Binomial}(8, 1-0.7).$ That is why the table only goes up to $0.5$.