8. An instructor passed out midterm reports after the first test. The report contained a homework grade and a test grade.   30%   of the class got unsatisfactory grades on both the homework and the test.   60%   got satisfactory grades on the homework, and   50%   got satisfactory grades on the test.

• a) What percent of the class got satisfactory grades on both the test and the homework?
• b) What percent of the students who got satisfactory grades on the homework got satisfactory grades on the test?
• c) Are getting satisfactory grades on the test and the homework independent events?

a) If the experiment is checking on student's grades, let   H   be the event that the student got a satisfactory grade on the homework, and let   T   be the event that the student got a satisfactory grade on the test. Let us make up a Venn diagram.

then since   30%   of the homework and the tests, we put a   .3   outside of the union. This tells us that

Pr(TH) = .7

When working with these Venn diagrams, it works best to start with the smallest set,   TH   and work out. Diabolically, TH   is the one thing we do not know. So we use the formula

Pr(TH) = Pr(T) + Pr(H) - Pr(TH)

We can fill in numbers

.7 = .5 + .6 - Pr(TH)

and solve for   Pr(TH).

Pr(TH) = .7 - .5 - .6 = .4

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b) We can now fill in the Venn diagram.

The question is, what percent of the students who got satisfactory grades on the homework got satisfactory grades on the test? We can use the formula for conditional probability.

which we can now see comes out to be

.4/.6 = 2/3

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c) In order for them to be independent events we would have to have

Pr(T|H) = Pr(T)

that is, the conditional probability would have to be the same as the original probability, but   2/3   is not equal to   .5.   In this case, the students' chances of doing well on the test improve if they do the homework.

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