Voter Fraud

The probability of any given arrangement of these 10 events is $(\frac{1}{2})^{10} = \frac{1}{1024}$ .

We must then multiply this probability by the number of combinations of 7 Monika votes and 3 Alfred votes, which is $\frac{10!}{7!3!} = 120$ .

$\frac{120}{1024}$ reduces to $\frac{15}{128}$ and 15 + 128 = 143.

The votes are distributed binomially as there are two candidates with a probability of success/failure $p$ (0.5 in this case) and a fixed number of trials $n$ .

Let $X =$ the number of votes for each candidate:

$X \sim B(10, 0.5)$

$P(X=3) = P(X=7) = \binom{10}{3}\cdot 0.5^{10} = \frac{15}{128}$

So $a + b = 143$

The machine will have counted the votes correctly if Monika receives $7$ out of the $10$ votes (consequently, Alfred receives $3$ ). The probability of this happening is $\dfrac{\binom{10}{7}}{2^{10}} = \dfrac{15}{128}.$

Since there are 10 votes and the vote is given to each candidate with probability 1/2, the total possible number of combination of votes is 2^10=1024 And how many ways are there for Alfred to get 3 votes= Using combination formula , 10!/3!7!=120 Since when Alfred get 3 votes, Monika will get 7 votes hence even though the ways for Monika to get 7 votes is also 120, we don't need to double count it. The probability of the correct number of votes is 120/1024=15/128 Since 15 and 128 are coprime, Thus a+b=15+128=143

There are $2^{10} = 1024$ different possible ways the machine could have assigned the votes to the students. Of these, $\binom{10}{7} = 120$ would give the correct number of votes to the two candidates. Thus, the probability that the machine gave each student the correct number of votes is $\frac{120}{1024} = \frac{15}{128}$ . Therefore, $a + b= 15 + 128 = 143$ .