Answer:
The required probability is 0.927
Step-by-step explanation:
Consider the provided information.
Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.
That means 95% of students didn't enrolled in SAT prep course.
Let P(SAT) represents the enrolled in SAT prep course.
P(SAT)=0.05 and P(not SAT) = 0.95
30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.
P(F) represents the first choice college.
The probability he didn't take an SAT prep course is:
[tex]P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}[/tex]
Substitute the respective values.
[tex]P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}[/tex]
[tex]P[\text{not SAT} |P(F)]\approx0.927[/tex]
Hence, the required probability is 0.927
