Question

at metro, 18% of all students play both football and basketball, and 32% of all students play football. we want to find the probability that a student plays basketball given that the student plays football.
a. let’s first define events a and b.
event a = student plays \boxed{quad}, event b = student plays \boxed{quad}
b. identify probabilities for each event. p(a) = \boxed{quad}, p(a ∩ b) = \boxed{quad}
write your answer as a decimal.
c. using the formula below, what is the probability that a student plays basketball given that the student plays football?
p(b|a) = \frac{p(b ∩ a)}{p(a)} p(b|a) = \boxed{quad}
round your answer as a decimal to the nearest hundredth.

Explanation:

Part a: Define Events

We need to define events \( A \) and \( B \) based on the problem. The problem is about finding the probability that a student plays basketball given that the student plays football. So, we can define:

• Event \( A \): student plays football (since we are given the probability of playing football and using it as the condition)
• Event \( B \): student plays basketball (the event we are interested in given the condition)

Part b: Identify Probabilities

• \( P(A) \) is the probability that a student plays football. From the problem, 32% of all students play football. To convert a percentage to a decimal, we divide by 100. So, \( P(A) = \frac{32}{100} = 0.32 \).
• \( P(A \cap B) \) is the probability that a student plays both football and basketball. From the problem, 18% of all students play both. Converting to a decimal, we get \( P(A \cap B) = \frac{18}{100} = 0.18 \).

Part c: Calculate Conditional Probability

We use the formula for conditional probability: \( P(B|A) = \frac{P(B \cap A)}{P(A)} \). We know that \( P(B \cap A) = P(A \cap B) = 0.18 \) (since intersection is commutative, \( A \cap B = B \cap A \)) and \( P(A) = 0.32 \). Plugging these values into the formula:

$$ P(B|A) = \frac{0.18}{0.32} = 0.5625 $$

Rounding to the nearest hundredth, we look at the thousandth place (2). Since 2 < 5, we round down. So, \( P(B|A) \approx 0.56 \).

Answer:

s:

Part a:

Event \( A \) = student plays football, Event \( B \) = student plays basketball

Part b:

\( P(A) = 0.32 \), \( P(A \cap B) = 0.18 \)

Part c:

\( P(B|A) = 0.56 \) (rounded to the nearest hundredth)