Using conditional probability, it is found that there is a 0.2273 = 22.73% probability that a student took AP Chemistry, given they did not get into their first-choice college.
Conditional Probability
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Did not get into their first-choice college.
- Event B: Took AP chemistry.
According to the chain given, the percentages associated with not getting into their first-choice college are:
Hence:
[tex]P(A) = 0.1375 + 0.1575 + 0.24 + 0.07 = 0.605[/tex]
The probability of both not getting into their first-choice college and taking chemistry is 0.1375, hence [tex]P(A \cap B) = 0.1375[/tex]
Then, the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1375}{0.605} = 0.2273[/tex]
0.2273 = 22.73% probability that a student took AP Chemistry, given they did not get into their first-choice college.
To learn more about conditional probability, you can take a look at https://brainly.com/question/14398287