What is the probability that a randomly selected freshman will enroll exactly one GE course and that course is not ENG10?

Question

In a survey on college freshmen, the following information were gathered:

1. All freshmen will enroll at least one general education (GE) course.
2. 40% of the freshmen will enroll more than one GE course.
3. 50% of the freshmen will enroll ENG10.
4. 70% of the freshmen who will enroll more than one GE will enroll ENG10.

Answer 1

Look up conditional probabilities:

Probability of enrolling in one GE = P(GE)= 0.4

Probability of enrolling in more than one GE = P(mGE)= 1-0.4 = 0.6

Prob of enrolling in English = P(Eng) = 0.5

Prob of not enrolling in English = P(nEng) = 1-0.5

∩ = and

order does not matter

Want: P(GE ∩ nEng)

Prob of enrolling in enrolling in english, given the student enrolls in more than one GE = P(Eng given mGE) = 0.7

P(Eng given mGE) = P(Eng ∩ mGE) / P(mGE) = 0.7

P(Eng ∩ mGE) = 0.7 * P(mGE) = 0.7 * 0.6 = 0.42

P(Eng) = P(Eng ∩ mGE) + P(Eng ∩ GE)

0.5 = 0.42 + P(Eng ∩ GE)

0.5-0.42 = P(Eng ∩ GE) = 0.08

P(GE) = P(Eng ∩ GE) + P(nEng ∩ GE)

0.4 = 0.08 + P(nEng ∩ GE)

0.4-0.08 = P(nEng ∩ GE) = 0.32