Question

question 1 of 39
at a high school, 18% of the students play football and 6% of the students play football and baseball. what is the probability that a student plays baseball given that he plays football?
a. 10.8%
b. 3%
c. 1.1%
d. 33.3%

Explanation:

Step1: Recall the conditional probability formula

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Let \(A\) be the event that a student plays baseball and \(B\) be the event that a student plays football.

Step2: Substitute the given values into the formula

We are given that \(P(B) = 0.18\) (18% play football) and \(P(A\cap B)=0.06\) (6% play both football and baseball). Then \(P(A|B)=\frac{0.06}{0.18}\).

Step3: Simplify the fraction

\(\frac{0.06}{0.18}=\frac{6}{18}=\frac{1}{3}\approx0.333 = 33.3\%\)

Answer:

D. 33.3%