Question

A) A survey of 600 undergraduate students in the college of engineering and information technology at an American University revealed the following regarding the gender and majors of the students: Major Gender CS GIS IT Total Male 120 150 80 350 Female 100 90 60 250 Total 220 240 140 600 1) What is the probability of selecting a male student? 2) What is the probability of selecting a IT majors? 3) What is the probability of selecting an mail student or IT majors? 4) What is the probability of selecting a CS or a GIS ? (2 points) 5) What is the probability of selecting a female and a non-GIS ? 6) Are "being a male" and "major IT" independent? Why? 7) What is the probability of selecting an GIS, given that the person selected is a male? B) Epidemiologists claim that the probability of breast cancer among Caucasian women in their mid-50s is 0.005. An established test identified people who had breast cancer and those that were healthy. A new mammography test in clinical trials has a probability of 0.85 for detecting cancer correctly. In women without breast cancer, it has a chance of 0.925 for a negative result. If a 55-year-old Caucasian woman tests positive for breast cancer, what is the probability that she, in fact, has breast cancer?

B) Epidemiologists claim that the probability of breast cancer among
Caucasian women in their mid-50s is 0.005. An established test identified
people who had breast cancer and those that were healthy. A new
mammography test in clinical trials has a probability of 0.85 for detecting
cancer correctly. In women without breast cancer, it has a chance of 0.925
for a negative result. If a 55-year-old Caucasian woman tests positive for
breast cancer, what is the probability that she, in fact, has breast cancer?

Answer 1

(A)

(1)
P(Male) = 350/600= 0.5833

(2)
P(IT) = 140/600 = 0.2333

(3)
P(Male OR IT) = P(Male) +P(IT) - P(Male AND IT)

                    = 350/600 + 140/600 - 80/600

                   = 410/600

                  = 0.6833

(4)

P(CS OR GIS)= P(CS) + P(GIS)

                       = 220/600 + 240/600

                         = 460/600

                         = 0.7667

(5)

P(Female OR Non-GIS) = P(Female) + P(Non-GIS)- P(Female AND Non-GIS)

                                = 250/600 + 360/600 - 160/600

                               = 450/600

                                  = 0.75

(6)
P(Male) = 350/600 = 0.5833

P(IT) = 140/600 = 0.2333

P(Male) X P(IT) = 0.5833 X 0.2333 = 0.1361

P(Male AND IT) = 80/600 = 0.1333

Since P(Male) X P(IT) = 0.1361 $\neq$ P(Male AND IT) = 0.1333, the events "being a male" and "major IT" are not independent.

(7)

P(GIS/ Male) = P(GIS AND Male)/ P(Male)

                 = 150/350

                 = 0.4286

(B)

From the given data, the following Table is calculated:

	Has breast cancer	Does not have breast cancer	Total
Test positive	0.005 X 0.85 = 0.00425	0.074625	0.078875
Test negative	0.00075	0.995 X 0.925 = 0.920375	0.921125
Total	0.005	0.995	1.00

P(Has breast cancer/ Test positive) = P(Has breast cancer AND Test positive) /P(Test positive)

                                                  =0.00425/ 0.078875

                                                = 0.0539

So,

Answer is:

0.0539