Question

suppose that 3% of all adults suffer from diabetes and that 32% of all adults are obese. suppose also that 2% of all adults both are obese and suffer from diabetes. (a) find the probability that a randomly chosen adult is obese, given that he or she suffers from diabetes. round your answer to the nearest hundredth. (b) find the probability that a randomly chosen adult who is obese also suffers from diabetes. round your answer to the nearest hundredth.

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of event \( A \) given event \( B \), \( P(A \cap B) \) is the probability of both \( A \) and \( B \) occurring, and \( P(B) \) is the probability of event \( B \) occurring.

Step2: Solve Part (a)

Let \( A \) be the event that an adult is obese, and \( B \) be the event that an adult suffers from diabetes. We know \( P(B) = 0.03 \) (3% of adults have diabetes), and \( P(A \cap B) = 0.02 \) (2% of adults are both obese and have diabetes). Using the conditional probability formula:
\( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.02}{0.03} \approx 0.67 \) (rounded to the nearest hundredth).

Step3: Solve Part (b)

Now, we want \( P(B|A) \), the probability an adult has diabetes given they are obese. We know \( P(A) = 0.32 \) (32% of adults are obese), and \( P(A \cap B) = 0.02 \). Using the conditional probability formula:
\( P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.32} = 0.0625 \approx 0.06 \) (rounded to the nearest hundredth).

Answer:

(a) \( 0.67 \)
(b) \( 0.06 \)