Question

Before graduating this year, a senior homeroom was given a survey. Of those surveyed, 24% felt they learned better at home. Of this group, 80% said they plan on taking an online course in college. Of the students who felt they did not learn better at home, 40% said they plan on taking an online course in college

Part A
What is the probability a person who does not plan on taking an online course felt they learned better at home?

A : 2/21
B : 24/125
C : 38/125
D : 19/31
E :None of these

Part B
What is the probability a person who does plan on taking an online course felt they did not learn better at home?
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A : 2/21
B : 24/125
C : 38/125
D : 19/31
E : None of these

Answer 1

(A) The correct option is (A).

(B) The correct option is (E).

Explanation:

The events can be defined as follows:

X = students felt they learned better at home

Y = students plan on taking an online course in college

The information provided is:

P (X) = 0.24

P (Y|X) = 0.80

P (Y|X') = 0.40

`P(Y'|X)=1-P(Y|X)\\=1-0.80\\=0.20`

`P(Y'|X')=1-P(Y|X')\\=1-0.40\\=0.60`

The Bayes' theorem states that the conditional probability of an event E
`_(i)` given that another event A has already occurred is:

`P(E_(i)|A)=\fracE_(i))P(E_(i)){\sum P(A}`

(A)

Compute the probability a person who does not plan on taking an online course felt they learned better at home as follows:

Use the Bayes' theorem.

`P(X|Y')=(P(Y'|X)P(X))/(P(Y'|X)P(X)+P(Y'|X')P(X'))`

`=(0.20* 0.24)/((0.20* 0.24)+(0.60* 0.76))\\\\=0.09524\\\\\approx 0.095`

Thus, the probability a person who does not plan on taking an online course felt they learned better at home is 0.095 or 2/21.

(B)

Compute the probability a person who does plan on taking an online course felt they did not learn better at home as follows:

`P(X'|Y')=1-P(X|Y')\\=1-0.095\\=0.905`

Thus, the probability a person who does plan on taking an online course felt they did not learn better at home is 0.905.