practice_problems/problems/unit-01/D-unions-intersections-and-complements/1

[giscus[bot]]

practice_problems/problems/unit-01/D-unions-intersections-and-complements/1

https://mids-w203.github.io/practice_problems/problems/unit-01/D-unions-intersections-and-complements/1.html

[IreneNS]

Maybe the question should be changed to ask 'how many of the students have at least one of these bad habits' instead of asking for the fraction, given we are not given the total number of the students

[IreneNS]

or rephrase the question as percentage as it seems to have suggested in the solution.

[william061490]

Right, 70 is the number of students, not fraction.

[shanis345]

Total number of students should be informed as 100 in order to make this solution valid. Else, we can only know the number of students that meet the condition. Not probability.

[dcaponi]

I agree. I looked up the Berkeley student population to do this and couldn't get the math to math :)

[bryan-md]

This one is very odd. The total number of students is not given. Nor is it explicit if the 15 that do both are in either of the other two groups.

With the fractions used, wouldn't that mean there are 100 students, and the 15 are not included in the other groups, making these 3 sets disjoint? in that case the intersection would be 0, and the fraction would be 100/100.

If the 15 ARE included in the other sets, and without the total students given, then the assumption is that the total group size is 70. Which would make the fraction 70/70, because again everyone has at least 1 bad habit

I suppose another scenario where 15 students are the intersection (.15). This would make P(A)+P(C) < 1
P(C\A) = .10
P(A\C) = .45
And would mean there is a third set comoliment to A, C, and the intersection that = .3
Which would make the fraction 70/100.