Probability that ensemble model is correct based on accuracies of its classifiers I'm trying to understand what I did wrong when trying to answer this question.  The exact question is:

Assume that we have 3 trained prediction models, and each model outputs either
-1 or 1. We then tested the accuracies of these models and obtained the following outcomes:





Model
Accuracy




m1
0.60


m2
0.55


m3
0.45





Let M be the ensemble model that outputs a plurality vote of these three models. If we assume that the errors of the models m1, m2, and m3 are independent, what is the probability that M(x) would be correct on a test instance x?

I thought that because this was a plurality vote, and the classification errors are independent of each other, I could simply take the weighted average of the accuracy of the three classifiers:
$
\begin{align*}
P(X) &amp;= \sum_{all\ models\ M_j} P(C_i|x,M_j)P(M_j) \\
 &amp;= \frac{1}{L} \sum_{all\ models\ M_j} P(C_i|x,M_j) \\
 &amp;= \frac{1}{3}(0.60+0.55+0.45)\\
 &amp;= 0.53
\end{align*}
$
But I was told that this is incorrect (with no context as to why).
Can someone explain why this is incorrect?  If this is a plurality vote (which to me assumes that the votes of each classifier are equal), why can I not simply take the weighted average?
It's not the actual data, it's the probabilities. So you should consider all the scenarios of voting.
For the Ensemble to be correct,
Either any two or all the three should be correct
=$[m_1*m_2*(1- m_3) + m_1*(1-m_2)*m_3) + (1-m_1)*m_2*m_3] + [m_1*m_2*m_3]$
= [0.6*0.55*(1-0.45) + 0.6*(1-0.55)*0.45 + (1-0.6)*0.55*0.45] + 0.6*0.55*0.45
= 0.5505