given that There were 2 contenders for mayor, and contender A get 2656 votes and contender B get 2594 votes. If after the elections found that 136 from all the electors were "invalid voters".
How can I calculate the probability for that A is not the really winner?
I thought to define this sample space:
{ $ (\alpha,\beta) $ | $ \alpha+\beta=136$ }
And in addition, the Event C that represent that A is not the winner:
C = { $ (k,m) $ | $ m+k=136$ , $2656-k \leq 2594-m$ }
Namely:
C = { $ (99,37) , (100,36) , ... , (136,0) $ }
How can I continue from this point for calculate $P(C)$ ?
We need to make an important assumption here, that the probability that a particular invalid voter voted for A or B is $1/2$, independent of all others.
So, for any $0≤n≤136$, the probability that the $n$ of the invalid voters voted for A (whilst the other $136-n$ voted for B) is the binomial
$$C^{136}_n\Bigl(\frac 12 \Bigr)^n\Bigl(\frac 12 \Bigr)^{136-n}=C^{136}_n\Bigl(\frac 12 \Bigr)^{136}$$
Thus,
$$\Bbb P[(99,37)]=C^{136}_{99}\Bigl(\frac 12 \Bigr)^{136}$$ $$\Bbb P[(100,36)]=C^{136}_{100}\Bigl(\frac 12 \Bigr)^{136}$$
and so on, hence
$$\Bbb P(C)=\sum_{n=99}^{136}C^{136}_n\Bigl(\frac 12 \Bigr)^{136}=\Bigl(\frac 12 \Bigr)^{136}\sum_{n=99}^{136}C^{136}_n$$