Question

question
a npr/robert wood johnson foundation/harvard t.h. chan school of public health poll asked adults whether they participate in a sport. given that the probability that a randomly chosen adult is a woman is 0.51, and the probability that a randomly chosen adult is a woman and plays a sport is 0.08, what is the probability of an adult playing a sport given that the adult is a woman?
round your answer to two decimal places.
provide your answer below:

Explanation:

Step1: Recall the formula for conditional probability

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: Identify the events and their probabilities

Let \( A \) be the event that an adult plays a sport, and \( B \) be the event that an adult is a woman. We are given \( P(B) = 0.51 \) (probability that an adult is a woman) and \( P(A \cap B) = 0.08 \) (probability that an adult is a woman and plays a sport). We need to find \( P(A|B) \) (probability that an adult plays a sport given that the adult is a woman).

Step3: Apply the conditional probability formula

Substitute the given values into the formula: \( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.08}{0.51} \).

Step4: Calculate the result and round

Calculate \( \frac{0.08}{0.51} \approx 0.15686 \). Rounding to two decimal places gives \( 0.16 \).

Answer:

\( 0.16 \)