A professor knows that in her class 37% of the students have passed an Algebra course, 29% have passed an English course, and 21% have passed both courses. If one student is selected at random from this class, find the probability that:
37% passed algebra,
21% passed both, so
(37-21)=16% passed only algebra.
Similarly,
(29-21)=8% passed only English.
With these numbers, you should have no difficulties calculating the probabilities that you forgot to post.
To find the probability in this case, we need to use the concept of set theory and probability. Let's break down the information given:
- 37% of students have passed an Algebra course.
- 29% of students have passed an English course.
- 21% of students have passed both an Algebra and an English course.
We can represent these percentages using sets:
Let A represent the set of students who passed Algebra.
Let E represent the set of students who passed English.
From the given information, we can deduce the following:
- The probability of students passing Algebra, denoted P(A), is 0.37 (37%).
- The probability of students passing English, denoted P(E), is 0.29 (29%).
- The probability of students passing both Algebra and English, denoted P(A∩E), is 0.21 (21%).
Now, let's find the probability of the different scenarios we need to compute:
1. The probability that a student passed Algebra or English (P(A∪E)):
To find this probability, we need to calculate the union of sets A and E, which represents students passing either course or both. We can use the formula:
P(A∪E) = P(A) + P(E) - P(A∩E)
P(A∪E) = 0.37 + 0.29 - 0.21
= 0.45 or 45%
2. The probability that a student did not pass either Algebra or English (P((A∪E)') or P(not passing either course)):
To find this probability, we need to subtract the probability of passing Algebra or English from 1, as it represents the complement of passing either course.
P(not passing either course) = 1 - P(A∪E)
P(not passing either course) = 1 - 0.45
= 0.55 or 55%
3. The probability that a student passed Algebra but not English (P(A∩E')):
To find this probability, we need to subtract the probability of passing both courses (P(A∩E)) from the probability of passing Algebra (P(A)).
P(A∩E') = P(A) - P(A∩E)
P(A∩E') = 0.37 - 0.21
= 0.16 or 16%
4. The probability that a student passed English but not Algebra (P(E∩A')):
To find this probability, we need to subtract the probability of passing both courses (P(A∩E)) from the probability of passing English (P(E)).
P(E∩A') = P(E) - P(A∩E)
P(E∩A') = 0.29 - 0.21
= 0.08 or 8%
These are the probabilities we can compute from the given information using set theory and probability.