Bayes' theorem : lie-detector machine Assuming there is lie-detector which can find out 95% of all lies correctly and of all true statements classifies 98% as true. Now we know that only one person would lie among 300. If the detector says a person is lying, what is the probability that this person is lying?
Assuming that X = {person is lying}, D = {detector finds a lie}, then we need to know p(X|D) given that p(X) = 1/300. Now I'm stuck with the meaning of 95% and 98%,   is p(D|X) = 0.95*0.98?
 A: Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively.  let $\bar X$ and $\bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$\Pr[D \mid X] = 0.95, \quad \Pr[\bar D \mid \bar X] = 0.98, \quad \Pr[X] = \frac{1}{300}.$$  We want to compute $\Pr[X \mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied.  Bayes' rule gives
$$\Pr[X \mid D] = \frac{\Pr[D \mid X]\Pr[X]}{\Pr[D]},$$ the numerator of which is already known.  The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$\Pr[D] = \Pr[D \mid X]\Pr[X] + \Pr[D \mid \bar X]\Pr[\bar X].$$
The first term is the same as the numerator.  The second term is computed by noting $$\Pr[\bar X] = 1 - \Pr[X], \\ \Pr[D \mid \bar X] = 1 - \Pr[\bar D \mid \bar X].$$  Therefore, in terms of the given probabilities, we have
$$\Pr[X \mid D] = \frac{\Pr[D \mid X]\Pr[X]}{\Pr[D \mid X]\Pr[X] + (1 - \Pr[\bar D \mid \bar X])(1 - \Pr[X])}.$$
