Where P E i i - P E , m
$\square$
represents the chance of j errors among the m base level classifiers If the probability of making an E error is less than

$\mathbf{f}\cdot\dot{\mathbf{z}}$ . 
then the majority filter will make fewer errors than a single algorithm filter formed from one of the base level classifiers The probability of mistaking a mislabeled instance for a correctly labeled instance E
occurs when more than half of the base level classifiers classify the instance as the mislabeled class 2 Let P E i be the probability that a base level detector i makes an error of type Assuming that the errors are independent and that the probabilities of the base level E
classifiers making an E error are the same then the probability that the majority vote filter makes a type E  error is given by

$$\sum_{j\ \ m/2}^{j=m}P\ E\ _{i}\ ^{j}\ \ -\ P\ E\ _{i}\ ^{m-j}\left(\begin{array}{c}{{m}}\\ {{j}}\end{array}\right)$$
$$P\setminus E$$

Therefore a majority vote filter will make fewer E errors than a single algorithm filter if P E is less than When all base level classifiers E errors are made on the same subset of the instances  the probability that a majority vote classifier will make an E  error is identical to the probability that a single algorithm filter will make an error

## 3.3.2 Consensus Filters

For a consensus filter an E error occurs when all of the base level detectors fail to classify an instance correctly  Let P E i  be the probability that base level detector i makes an E
error and m be the number of base level classifiers then the general form of the probability of making an E  error is given by

$$P\setminus E$$
$P\ E\ _{1}\ P\ E\ _{2}\ |\ E\ _{1}\...P\ E\ _{n}\ |\ E\ _{1}\ \cap\ \cdot\ \cdot\ E\ _{m-1}$
If the base level detectors make errors on the same instances then the probability of an E
error is equal to the probability that a single base level detector makes an error P E ;
When the E errors of the base level detectors are independent then a consensus filter has a smaller probability of making an E error than each of its base level detectors and the probability of making an E  error is given  by

$$\prod_{i=1}^{m}P\setminus E\setminus i$$
$$P\setminus E$$

If the assumption of independence of the errors of the base level detectors holds  then we would expect a consensus filter to have a smaller probability of making an E  error than a single algorithm filter