Among 500 freshmen pursuing a business degree at a university, 313 are enrolled in an economics course, 205 are enrolled in a mathematics course, and 153 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in
(a) An economics and/or a mathematics course?
(b) Exactly one of these two courses?
(c) Neither an economics course nor a mathematics course?
I may have read this wrong, of course. After reading it again, I thought differently.
There are 153 students taking both economics and math. That means there are:
205 - 153 = 52
There are 52 students only taking math, but not economics.
313-153 = 160
So 160 students are taking economics without math.
So out of 500, we have:
160 only taking economics.
52 only taking math.
153 taking both.
That is 365 out of 500 taking at least one of the classes. How many aren't taking math or economics?
Now, it's just a matter of percentages.
So out of 500 students, we know what 671 of them are doing. I think there are many unregistered freshman taking classes at that college. ...
The numbers do not add up.
Half of a third of x equals a fourth of y plus a fifth of y. If x = 27, what is the value of y?
4 and 5
To find the probabilities in this problem, we need to use the principles of probability and set theory. Let's break down the problem step by step:
(a) An economics and/or a mathematics course:
To find the probability that a freshman is enrolled in either an economics or a mathematics course or both, we need to find the union of the two events: P(E or M).
To do this, we can use the formula for the union of two events: P(E or M) = P(E) + P(M) - P(E and M), where E represents the event of being enrolled in an economics course, M represents the event of being enrolled in a mathematics course, and E and M represents the event of being enrolled in both courses.
Given:
P(E) = 313 (enrolled in economics)
P(M) = 205 (enrolled in mathematics)
P(E and M) = 153 (enrolled in both economics and mathematics)
Using the formula, the probability that a freshman is enrolled in an economics and/or a mathematics course is:
P(E or M) = P(E) + P(M) - P(E and M)
P(E or M) = 313 + 205 - 153
P(E or M) = 365
Therefore, the probability that a freshman selected at random from this group is enrolled in an economics and/or a mathematics course is 365/500 or 0.73.
(b) Exactly one of these two courses:
To find the probability that a freshman is enrolled in exactly one of the two courses (economics or mathematics), we need to subtract the probability of being enrolled in both courses from the probability of being enrolled in either one of the courses.
This can be calculated as:
P(Exactly one course) = P(E or M) - P(E and M)
P(Exactly one course) = 365 - 153
P(Exactly one course) = 212
Therefore, the probability that a freshman selected at random from this group is enrolled in exactly one of the two courses is 212/500 or 0.424.
(c) Neither an economics course nor a mathematics course:
To find the probability that a freshman is not enrolled in an economics course and not enrolled in a mathematics course, we can subtract the probability of being enrolled in either one or both courses from 1 (the total probability).
This can be calculated as:
P(Neither E nor M) = 1 - P(E or M)
P(Neither E nor M) = 1 - 0.73
P(Neither E nor M) = 0.27
Therefore, the probability that a freshman selected at random from this group is neither enrolled in an economics course nor a mathematics course is 0.27 or 27%.