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3 classifiers. Each classifier has $0.7$ accuracy and makes its error independently.

How do I calculate the probability that the majority of three classifiers are wrong?

how I'm trying to solve it:

(3 choose 2) * 0.3 * 0.3 * 0.7 * 0.3 * 0.3 * 0.3

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  • $\begingroup$ What made you decide to use this particular formula? $\endgroup$ – Ben Grossmann Nov 22 '13 at 2:28
  • $\begingroup$ I think the majority is 3 choose 2. Then I multiply by what is left times the .7 accuracy. That came into my head when I saw the problem, I am not sure it is an actual formula. $\endgroup$ – CharlieK Nov 22 '13 at 2:56
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Hint: Majority means it could potentially be either $3$ or $2$. The probability of majority of the classifiers being wrong is the probability that either $2$ are wrong or all $3$ are wrong.
Example: Probability that $k$ out of $n$ independent trials of a random experiment are successful, with the probability of success being $p$ is ${n \choose k}p^k(1-p)^{n-k}$.

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  • $\begingroup$ I don't understand that formula. I agree with the n choose k part, but I don't understand the rest. $\endgroup$ – CharlieK Nov 22 '13 at 3:00
  • $\begingroup$ The rest is just an hint as to this particular experiment is equivalent to an experiment with a Binomial Distribution. Given that there are $k$ successes out of $n$ trials, the probability of such an occurrence is $p^k(1-p)^{n-k}$ since the trials are independent. This distribution just tells you as to how many of those events can be accounted for and that is ${n\choose k}$. And why the downvotes? $\endgroup$ – Sudarsan Nov 23 '13 at 18:51
  • $\begingroup$ $n = 2$ and $k = 3$ but what does $p = ?$ $\endgroup$ – CharlieK Nov 23 '13 at 19:33
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    $\begingroup$ $p$ is the probability of success/failure (depending upon the way you want to interpret) that you consider for the experiment. Here in our question the accuracy is $0.7$ and hence $p=0.7$ which is the probability of success and $1-p=0.3$ which is the probability of failure. $\endgroup$ – Sudarsan Nov 23 '13 at 19:37
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    $\begingroup$ The simplified form would be ${3\choose 2} 0.3^2 0.7 + {3\choose 3} 0.3^3$ and that is your answer. $\endgroup$ – Sudarsan Nov 23 '13 at 20:34
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A similar question has been asked/answered here, but with 17 classifiers. Simply use their solution and change from 17 to 3 to apply to here. To improve the accuracy, a majority vote is taken. What is the probability of an incorrect classification?

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  • $\begingroup$ mark as duplicate instead of answering $\endgroup$ – suomynonA Dec 6 '16 at 4:21

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