In a certain school, 17 percent of the students are enrolled in a psychology course, 28 percent are enrolled in a foreign language course, and 32 percent are enrolled in either a psychology course or a foreign language course or both. What is the probability that a student chosen at random from this school will be enrolled in both a foreign language course and a psychology course? A. 0.45 B. 0.32 C. 0.20 D. 0.13 E. 0.05

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Answer:

The required probability is 0.13

Step-by-step explanation:

Consider the provided information.

It is give that 17% of the student are enrolled in a psychology course.

Let P(A) is the student enrolled in psychology course.

Thus. P(A)=17%=0.17

28 % are enrolled in a foreign language course.

Let P(B) represents the student enrolled in foreign language course.

Thus. P(B)=28%=0.28

32 % are enrolled in either a psychology course or a foreign language course or both.

That means P(A∪B) = 32% = 0.32

We need to find the probability that a student chosen at random from this school will be enrolled in both a foreign language course and a psychology course.

That means we need to find P(A∩B).

P(A∪B) = P(A)+P(B)-P(A∩B)

0.32 = 0.17 + 0.28 - P(A∩B)

0.32 = 0.45 - P(A∩B)

P(A∩B) = 0.13

The required probability is 0.13

The required probability will be D. 0.13

Based on the information given,

  • Let P(A) is the student enrolled in psychology course. P(A)=17%=0.17
  • Let P(B) represents the students that are enrolled in foreign language course. P(B)=28%=0.28

Since we are told that 32% of the students are enrolled in either a psychology course or a foreign language course or both. This will be P(A∪B) = 32% = 0.32

Then, we need to find P(A∩B) and.this goes thus:

P(A∪B) = P(A)+P(B)-P(A∩B)

0.32 = 0.17 + 0.28 - P(A∩B)

Therefore, 0.32 = 0.45 - P(A∩B)

P(A∩B) = 0.45 - 0.32 = 0.13

Therefore, the required probability is 0.13

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