Question

4. At a college, 60 percent of the students are female and 35 percent of the students receive a grade of C. About 40 percent of the students are female and “not C” students. Use this contingency table.

Gender\Grade	C	Not C or C bar
Female (F)		0.40	0.60
Male (M)
	0.35

If a randomly selected student is a “not C” student, what is the probability the student  is a male student?

5. A and B are independent events. Moreover, P(A) = 0.8 and  P(B) = 0.6. Determine P(A  B), that is,  P(A or B)

Answer 1

(4)

From the given data, the following Table iscalculated.

	C	Not C	Total
Female (F)	0.20	0.40	0.60
Male (M)	0.15	0.25	0.40
Total	0.35	0.65	1.00

P(Male/ Not C) = P (Male AND Not C)/ P(Not C)

                      = 0.25/0.65 = 0.3846

So,

Answer is:

0.3846

(5)

P(A or B) = P(A) + P(B) - P(A) P(B)

Substituting values, we get:

P(A or B) = 0.8 + 0.6 - (0.8 X 0.6)

             = 0.92

So,

Answer is:

0.92