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.
[Math] the complement of conditional probabilities
bayes-theoremelementary-set-theoryprobability
Related Solutions
Let's use a frequency table of a deterministic cohort of light aircraft that disappear. Suppose there are $N = 100$ such aircraft. As $70\%$ of these are discovered, this means $70$ aircraft belong in the group $D$, indicating that they are subsequently discovered, and $30$ aircraft belong in the group $\bar D$, indicating they are not discovered.
Among the $70$ discovered aircraft, $60\%$ have an emergency locator, so $$D \cap L = (70)(0.6) = 42$$ where $L$ represents the event that an aircraft has an emergency locator. Thus there are $$D \cap \bar L = (70)(0.4) = 28$$ that were discovered but had no emergency locator.
Similarly, among the $30$ undiscovered aircraft, $$\bar D \cap \bar L = (30)(0.9) = 27$$ had no emergency locator; and $$\bar D \cap L = (30)(0.1) = 3$$ had an emergency locator.
We summarize the above in the following table:
$$\begin{array}{c|c|c|c} & L & \bar L & \\ \hline D & 42 & 28 & 70 \\ \hline \bar D & 3 & 27 & 30 \\ \hline & 45 & 55 & 100 \end{array}$$
Therefore, given that an aircraft has an emergency locator--that is to say, is one of the $45$ aircraft in column $L$--the number of discovered aircraft is $42$, thus the proportion of such aircraft is $42/45 \approx 0.933$.
In the language of probability, where $D$ and $L$ are events, we are given $$\Pr[D] = 0.7, \quad \Pr[L \mid D] = 0.6, \quad \Pr[\bar L \mid \bar D] = 0.9,$$ and we wish to compute $$\Pr[D \mid L] = \frac{\Pr[L \mid D]\Pr[D]}{\Pr[L]}.$$ Then $$\Pr[L] = \Pr[L \mid D]\Pr[D] + \Pr[L \mid \bar D]\Pr[\bar D] = (0.6)(0.7) + (1 - 0.9)(1 - 0.7) = 0.45,$$ and $$\Pr[D \mid L] = \frac{(0.6)(0.7)}{0.45} = \frac{42}{45} \approx 0.933,$$ as we found using the deterministic cohort above.
Now my only confusion is how the
P(False Negative Result)differs from theNegative Predictive Value.
Consider this probability tree:
p: disease prevalence and other (prior) risk factors
v: test sensitivity
f: test specificity
D: Diseased
H: Healthy
+: Positive test result
-: Negative test result
The negative predictive value (NPV) is $$P(H|-)=\frac{P(H-)}{P(H-)+P(D-)}=\frac{(1-p)f}{(1-p)f+p(1-v)};$$
the true negative rate (specificity) is $$P(-|H)=f;$$
P(true-negative result) is $$P(H-)=(1-p)f;$$
P(false-negative result) is $$P(D-)=p(1-v);$$
the false omission rate (complement of the NPV) is $$P(D|-)=\frac{P(D-)}{P(D-)+P(H-)}=\frac{p(1-v)}{p(1-v)+(1-p)f};$$
the false discovery rate (complement of the PPV) is $$P(H|+)=\dots.$$
$P(\text{False Negative Result}) = P(\text{Disease|Negative})$
“$P(\text{False Negative Result})$” ambiguously could mean either $P(\text{False-Negative Result})$ or $P(\text{Disease|Negative}).$
The former $(2.00\%)$ is the second outcome (of four possible outcomes) in the probability tree, whereas the latter $(2.29\%)$ is conditioned on the knowledge that only outcomes 2 & 4 are possible.
Would it be correct to assume the Negative Predictive Value is associated with a test, but the P(False Negative Result) is associated with an event?
No, each term above is associated with both a test and an event. Some of them are conditional probabilities, whereas the others are not working with a reduced sample space.
Read more here: the accuracy of a medical test.
Best Answer
$$P(E\mid D')=1-P(E'\mid D')$$ and $$P(E'\mid D)=1-P(E\mid D)$$ if that is what you mean by complement