Deanna L. answered 09/22/15
Tutor
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Electrical engineering major and music lover with MIT degree
So let's look at the facts and translate them into probability terms:
500 Freshman pursuing a business degree at a university, 336 are enrolled in an economics course, 226 are enrolled in a mathematics course, and 128 are enrolled in both an economics and a mathematics course.
P(U)=500/500=1. This to say the universe of students is 500, economics, mathematics, or neither.
P(A)=336/500. Let's say the event A is a student enrolled in an economics course.
P(B)=226/500. Now event B is a student enrolled in a mathematics course.
Now P(A∩B)=128/500. This means that there are 128 students where event A and B occur together.
P(AUB) is the probability a student is enrolled in economics or mathematics (and/or is same thing in this case). According to our formula: P(AUB)=P(A)+P(B)-P(A∩B)=(336+226-128)/500=434/500
P(A∩B') is the probability that someone is taking economics but not math. P(A'∩B) is probability someone is taking math but not economics. Imagine two circles intersecting in one place, where one circle is students taking math and the other circle is students taking economics. The intersection is where students are taking both classes. Then you can see that
P(A∩B')=P(A)-P(A∩B)=(336-128)/500=208/500
P(A'∩B)=P(B)-P(A∩B)=(226-128)/500=98/500
Now if we add these two cases together: P(A∩B')+P(A'∩B)=(208+98)/500=306/500
What about students taking neither math nor econ? Imagine drawing a box outside those two intersecting circles and figuring out how many of those 500 students fall outside of the circles. That would be P(U)-P(AUB), or (500-434)/500=66/500.
You'll have to punch all these numbers out to get the decimals but it's the process that counts here.
Hope that helps!