I am working with a problem that uses Bayes Theorem and conditional probabilities. I have the conditional probability that a plane has an emergency locator $(E)$ given that it was discovered $(D)$ which is $P(E\mid D)=0.60$. Now I am given that $P(E'\mid D')=0.90$, where a plane does not have a emergency locator given that it was not discovered. I wanted to know what the complement of $P(E'\mid D')$ would be. Is it $P(E\mid D)$ or $P(E\mid D')$? I am not sure whether or not to flip the $D$ in the conditional.
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$$P(E\mid D')=1P(E'\mid D')$$ and $$P(E'\mid D)=1P(E\mid D)$$ if that is what you mean by complement
Michael Hardy
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Henry
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Yes, could you explain why the $D'$ isn't affected by the flip? Is it because it is given? – Kot Sep 27 '14 at 23:21

Exactly that. Given $D'$, if $E'$ did not occur then $E$ did, but still within the context of $D'$. – Henry Sep 27 '14 at 23:33

@Henry could you please show how can we describe from a Set theory perspective the event $ED$, if it's possible in a relatively simple way. I think this could help to understand why the "flip" occur in that way, since, for me at least, it's still a little bit obscure. Thank you! – Turquoise Tilt Oct 15 '21 at 07:50

1@GodelSpassky "$E\mid D$" is not meaningful as an event in the original probability space. If you roll a fair die, then the outcome space is $\Omega=\{1,2,3,4,5,6\}$. Suppose $D$ is the event of nonsix roll and $E$ is the event of an even roll. Then $D=\{1,2,3,4,5\}$ and $P(D)=\frac56$ and $E=\{2,4,6\}$ and $P(E)=\frac36$; all events here have probabilities which are multiples of $\frac16$. But $P(E \mid D)=\frac{P(\{2,4\})}{P(\{1,2,3,4,5\})}=\frac25$. You might say that conditioning on $D$ means the outcome space becomes $\Omega_D=\{1,2,3,4,5\}$ and $E_D=\{2,4\}$. – Henry Oct 15 '21 at 08:24
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It will be $P(E\mid D')$ irrespective of whether or not to flip the $D$ in the conditional.
Michael Hardy
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RiLi
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