In a community, $a$ persons are pro-choice, $b$ $(b < a)$ are pro-life, and $n$ $(n > a – b)$ are undecided. Suppose that there will be a vote to determine the will of the majority with regard to legalizing abortion. If by then all of the undecided persons make up their minds, what is the probability that those who are pro-life will win? Assume that it is equally likely that an undecided person eventually votes pro-choice or pro-life.
Let $X$ be the number of people who go from undecided to pro-life. We want to find the probability that $$a+X>b+1-X$$ or $$n>X>\frac{b+1-a}{2}$$ I know that for a binomial random variable with parameters $n$ and $p$, $$P(X=x)=\binom{n}{x}p^x(1-p)^n-x$$ if $x=0,1,2,...,n$. However, I'm not sure how to proceed from here.
Let $X_i\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Ber}\left(\frac12\right)$ for $1\leqslant i\leqslant n$. The event we are interested in is $\left\{b + \sum_{i=1}^n X_i >a\right\}$, or $\left\{\sum_{i=1}^n X_i>a-b\right\}$. Now, since $\mathbb P(X_1=0)=\mathbb P(X_1=1)=\frac12$, it follows by induction that $\mathbb P\left(\sum_{i=1}^n X_n =k\right) = \binom nk \left(\frac12\right)^n$ for $0\leqslant k\leqslant n$. We compute the probability as follows: $$ \mathbb P\left(\sum_{i=1}^n X_i>a-b\right) = \left(\frac12\right)^n \sum_{k=a-b+1}^n \binom nk. $$