Questions: The table below displays results from experiments with polygraph instruments. Find t probability that the subject lied, given that the test yields a positive result. Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) 10 42 33 8 The probability is . (Round to three decimal places as needed.)

Find the probability that the subject lied, given that the test yields a positive result.

Define the relevant probabilities.

Let \( A \) be the event that the subject lied, and \( B \) be the event that the test result is positive. We need to find \( P(A \mid B) \), which can be calculated using the formula: \[ P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} \]

Calculate \( P(B \mid A) \), \( P(A) \), and \( P(B) \).

From the data:

• Positive test results when the subject lied: \( 8 \)
• Total subjects who lied: \( 42 + 8 = 50 \)
• Total positive test results: \( 8 + 33 = 41 \)
• Total subjects: \( 10 + 42 + 33 + 8 = 93 \)

Thus, \[ P(B \mid A) = \frac{8}{50}, \quad P(A) = \frac{50}{93}, \quad P(B) = \frac{41}{93} \]

Substitute the values into the formula.

Now substituting the values into the formula for conditional probability: \[ P(A \mid B) = \frac{\left(\frac{8}{50}\right) \cdot \left(\frac{50}{93}\right)}{\frac{41}{93}} = \frac{8}{41} \]

Calculate the final probability.

Calculating \( P(A \mid B) \): \[ P(A \mid B) = \frac{8}{41} \approx 0.195 \] Thus, the probability that the subject lied given a positive test result is approximately \( 0.195 \).

The probability that the subject lied given a positive test result is \( \boxed{0.195} \).

The probability that the subject lied given a positive test result is \( \boxed{0.195} \).