P(basketball) = [tex] 15 \% = \frac{15}{100}=0.15 [/tex]
P(baseball and basktball) =[tex] 4 \% = \frac{4}{100}=0.04 [/tex]
We have to find the probability that a student plays baseball given that he or she plays basketball.
This is the case of Conditional Probability, whose formula is given as follow:
[tex] P(B/A)=\frac{P(A and B)}{P(A)} [/tex]
Now,
[tex] P(Baseball/Basketball)=\frac{P(Baseball and Basketball)}{P(Basketball)} [/tex]
=[tex] \frac{0.04}{0.15}=0.27 [/tex]
Answer: The probability is P = 26.7%
Step-by-step explanation:
We know that 15% of the students play basketball, and 4% of the students play basketball and baseball.
Now, suppose that we define N as the total number of students, we will have that:
0.15*N play basketball.
0.04*N play basketball and also baseball.
then, the probability that a student plays baseball given that he plays basketball is equal to the quotient between the number of players that play baseball and basketball and the number of students that play basketball:
P = 0.04*N/0.15*N = 0.04/0.15 = 0.267
if we multiply this number by 100% we have the percentage form:
P = 0.267*100% = 26.7%
