Question

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students. You overhear a high school student say he got into the college he wanted. What is the probability he didn't take an SAT prep course?

Answer 1

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:

$P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}$

Substitute the respective values.

$P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}$

$P[\text{not SAT} |P(F)]\approx0.927$

Hence, the required probability is 0.927