Question

In an ethnic group, 30% of the adult male population is known to have heart disease. A test indicates high cholesterol level in 80% of adult males with heart disease. But the test also indicates high cholesterol levels in 10% of the adult males with no heart disease. Then the probability (round off to 2 decimal places) that a randomly selected adult male from this population does not have heart disease given that the test indicates high cholesterol level equals .............

Hint:

Bayes' Theorem is useful when you need to update the probability of an event based on new evidence. Always ensure that you have the correct conditional probabilities and total probability to use in the formula.

Answer 1

Correct Answer: 0.2

Step 1: Define the events.
Let \( A \) be the event that the adult male has heart disease, and \( B \) be the event that the test indicates high cholesterol. We are given: \[ P(A) = 0.3, \quad P(B \mid A) = 0.8, \quad P(B \mid A^c) = 0.1. \] Step 2: Use Bayes' Theorem.
We want to calculate \( P(A^c \mid B) \), the probability that the adult male does not have heart disease given that the test indicates high cholesterol. By Bayes' Theorem: \[ P(A^c \mid B) = \frac{P(B \mid A^c) P(A^c)}{P(B)}. \] Step 3: Compute \( P(B) \).
The total probability \( P(B) \) is: \[ P(B) = P(B \mid A) P(A) + P(B \mid A^c) P(A^c) = 0.8 \times 0.3 + 0.1 \times 0.7 = 0.24 + 0.07 = 0.31. \] Step 4: Calculate \( P(A^c \mid B) \).
Now we can compute: \[ P(A^c \mid B) = \frac{0.1 \times 0.7}{0.31} = \frac{0.07}{0.31} \approx 0.23. \] Final Answer: \[ \boxed{0.23}. \]