A lie detector can tell if a person is lying 85 % of the time. However it incorrectly says a person telling the truth is lying 5 % of the time. In a certain grade school class 4 % lie on a regular basis. If the machine says the person is lying, what is the probability the person actually lied?

Let's denote the events:
- Event A: The person is actually lying.
- Event B: The lie detector indicates the person is lying.

We are given:
- P(B|A) = 0.85 (the lie detector correctly identifies a lie),
- P(B|\neg A) = 0.05 (the lie detector incorrectly identifies the truth as a lie),
- P(A) = 0.04 (the probability a person lies on a regular basis).

We need to find P(A|B) , the probability the person actually lied given that the machine says the person is lying.

We will use Bayes' Theorem:

P(A|B) = \frac{P(B|A) P(A)}{P(B)}

We can calculate P(B) with the Law of Total Probability:

P(B) = P(B|A)P(A) + P(B|\neg A)P(\neg A)

P(\neg A) = 1 - P(A) = 1 - 0.04 = 0.96

Plugging in the values, we get:

P(B) = 0.85 \times 0.04 + 0.05 \times 0.96

P(B) = 0.034 + 0.048 = 0.082

Now, we can find P(A|B) :

P(A|B) = \frac{0.85 \times 0.04}{0.082}

P(A|B) = \frac{0.034}{0.082}

P(A|B) = 0.4146

So, the probability that the person actually lied given that the lie detector says the person is lying is approximately 41.46\% .

\boxed{P(A|B) \approx 0.4146}