Question

The college suspects their star player James Strongarm has used an illegal performance-enhancing drug. Currently, the probability that James is a drug user is 0.37. They consider barring him from athletics. The real "monetary" values of the outcomes are difficult to assess in this case and, therefore, the value of each outcome is expressed in a funny money unit: Galleon (from Harry Potter Movies). The following are the payoffs for the potential outcomes:

Correctly identifying a drug user and barring this person from athletics: 40 Galleon

Falsely accusing a nonuser and barring this person from athletics: -60 Galleon

Not identifying a drug user and allowing this person to participate in athletics: -30 Galleon

Not barring a nonuser: 0 Galleon

Before they make the decision, they can ask for a drug test, which costs -1 Galleon per single athlete . In the drug testing, assume there are two possible test results: positive and negative. For a drug user, the probabilities of these outcomes are 0.95 and 0.05, respectively. For a nonuser, they are 0.01 and 0.99, respectively. Assume that 2% of all college athletes are drug users.

(When you use the probabilities from the flipped table, use cell references in the decision tree instead of typing the probabilities manually. All choices in the following questions are based on the decision tree built using this practice.)

Question 5 (1 point)

What is the EMV of making the decision without the test?

A -58.6

B -24

C -12.1

D -11.1

Question 6 (1 point)

If the first decision is to do the test or not and the second decision is to bar or not, what is the EMV of the whole tree? (Due to the rounding error, your answer may not match perfectly with the correct choice. Pick the choice closest to your answer.)

A -0.858

B -0.269

C   0.142

D 2.678

Question 7 (1 point)

What is the best decision alternative?

A Forgo the test and bar him.

B Forgo the test and do not bar him.

C Take the test. Bar him if the result is positive; do not bar him if the result is negative.

D Take the test, and do not bar him.

Question 8 (1 point)

If the test costs -10 Galleon, what is the best decision alternative?

A Forgo the test and bar him.

B Forgo the test and do not bar him.

C Take the test. Bar him if the result is positive; do not bar him if the result is negative.

D Take the test, and do not bar him.

Answer 1

SOLUTION:-

Calculate conditional probabilities

Notations P Drug user = Non user N = Test positive Test negative TP TN - Given Calculate Conditional Probabilites using Bayes' Theorem P(TP P)*P(P) + P(TP | N)*P(N) P(P) P(TP) = 02*0.95+0.01 0.9 0.02 0.0288 = - = P(TN | P)*P(P) P(TN |N)*P(N) P(TN) P(N) = .05*0.02+0.99*0.9 0.98 0.9712 P (TP P) P (TN P) = 0.95*0.02/0.0288 = 0.6597 P(P TP) P(NI TP) P(TP P)*P(P)/P(TP) P(TPIN)*P(N)/P(TP) =0.95 0.01*0.98/0.0288 0.05 0.3403 P(TN N) P (TP N) P(PITN) P(NITN) P(TN P)*P(P)/P(TN) P(TN N)*P(N)/P(TN) 0.05*0.02/0.9712 = 0.0010 0.99 0.99*0.98/0.9712 = 0.01 = 0.9990

Decision tree is as follows:

User 0.37 40 -23 Forego test and bar him B Non-user -60 0.63 -30 User 0.37 Forego test and do not bar -11.1 Non-user 0.63 0 Max EV User 0.6597 40 Bar him 0.142 5.972 Non-user 0.3403 -60 0.0288 Positive 5.972 User 0.6597 -30 Do not bar -19.792 Non-user 0.3403 Take the test 0.142 D User 0.0010 40 Bar him -59.897 Non-user 0.9990 -60 Negative 0.9712 -0.0309 User 0.0010 -30 Do not bar -0.0309, Non-user 0.9990 0

EV of node B = .37*40+.63*(-60) = -23

EV of node C = .37*(-30)+.63*0 = -11.1

EV of node G = .6597*40+.3403*(-60) = 5.972

EV of node H = .6597*(-30)+.3403*0 = -19.792

EV of node J = .0010*40+.9990*(-60) = -59.897

EV of node K = .0010*(-30)+.9990*0 = -0.0309

EV of node D = .0288*MAX(5.972, -19.792) + .9712*MAX(-59.897, -.0309) = 0.142

EV of node D is the maximum.

Therefore best decision strategy is:  

C) Take the test, Bar him if the result is positive, do not bar him if the result is negative

EV of this decision strategy = 0.142

EMV of the tree without the test = MAX(-23, -11.1) = -11.1

Calculate conditional probabilities

Notations P Drug user = Non user N = Test positive Test negative TP TN - Given Calculate Conditional Probabilites using Bayes' Theorem P(TP P)*P(P) + P(TP | N)*P(N) P(P) P(TP) = 02*0.95+0.01 0.9 0.02 0.0288 = - = P(TN | P)*P(P) P(TN |N)*P(N) P(TN) P(N) = .05*0.02+0.99*0.9 0.98 0.9712 P (TP P) P (TN P) = 0.95*0.02/0.0288 = 0.6597 P(P TP) P(NI TP) P(TP P)*P(P)/P(TP) P(TPIN)*P(N)/P(TP) =0.95 0.01*0.98/0.0288 0.05 0.3403 P(TN N) P (TP N) P(PITN) P(NITN) P(TN P)*P(P)/P(TN) P(TN N)*P(N)/P(TN) 0.05*0.02/0.9712 = 0.0010 0.99 0.99*0.98/0.9712 = 0.01 = 0.9990

Decision tree is as follows:

User 0.37 40 -23 Forego test and bar him B Non-user -60 0.63 -30 User 0.37 Forego test and do not bar -11.1 Non-user 0.63 0 Max EV User 0.6597 40 Bar him 0.142 5.972 Non-user 0.3403 -60 0.0288 Positive 5.972 User 0.6597 -30 Do not bar -19.792 Non-user 0.3403 Take the test 0.142 D User 0.0010 40 Bar him -59.897 Non-user 0.9990 -60 Negative 0.9712 -0.0309 User 0.0010 -30 Do not bar -0.0309, Non-user 0.9990 0

EV of node B = .37*40+.63*(-60) = -23

EV of node C = .37*(-30)+.63*0 = -11.1

EV of node G = .6597*40+.3403*(-60) = 5.972

EV of node H = .6597*(-30)+.3403*0 = -19.792

EV of node J = .0010*40+.9990*(-60) = -59.897

EV of node K = .0010*(-30)+.9990*0 = -0.0309

EV of node D = .0288*MAX(5.972, -19.792) + .9712*MAX(-59.897, -.0309) = 0.142

EMV of the whole tree = MAX(-23, -11.1, 0.142) = 0.142

EV of node D is the maximum.

Therefore best decision strategy is:  

C) Take the test, Bar him if the result is positive, do not bar him if the result is negative

EMV of EV of this decision strategy = 0.142

Calculate conditional probabilities

Notations P Drug user = Non user N = Test positive Test negative TP TN - Given Calculate Conditional Probabilites using Bayes' Theorem P(TP P)*P(P) + P(TP | N)*P(N) P(P) P(TP) = 02*0.95+0.01 0.9 0.02 0.0288 = - = P(TN | P)*P(P) P(TN |N)*P(N) P(TN) P(N) = .05*0.02+0.99*0.9 0.98 0.9712 P (TP P) P (TN P) = 0.95*0.02/0.0288 = 0.6597 P(P TP) P(NI TP) P(TP P)*P(P)/P(TP) P(TPIN)*P(N)/P(TP) =0.95 0.01*0.98/0.0288 0.05 0.3403 P(TN N) P (TP N) P(PITN) P(NITN) P(TN P)*P(P)/P(TN) P(TN N)*P(N)/P(TN) 0.05*0.02/0.9712 = 0.0010 0.99 0.99*0.98/0.9712 = 0.01 = 0.9990

Decision tree is as follows:

User 0.37 40 -23 Forego test and bar him B Non-user -60 0.63 -30 User 0.37 Forego test and do not bar -11.1 Non-user 0.63 0 Max EV User 0.6597 40 Bar him 0.142 5.972 Non-user 0.3403 -60 0.0288 Positive 5.972 User 0.6597 -30 Do not bar -19.792 Non-user 0.3403 Take the test 0.142 D User 0.0010 40 Bar him -59.897 Non-user 0.9990 -60 Negative 0.9712 -0.0309 User 0.0010 -30 Do not bar -0.0309, Non-user 0.9990 0

EV of node B = .37*40+.63*(-60) = -23

EV of node C = .37*(-30)+.63*0 = -11.1

EV of node G = .6597*40+.3403*(-60) = 5.972

EV of node H = .6597*(-30)+.3403*0 = -19.792

EV of node J = .0010*40+.9990*(-60) = -59.897

EV of node K = .0010*(-30)+.9990*0 = -0.0309

EV of node D = .0288*MAX(5.972, -19.792) + .9712*MAX(-59.897, -.0309) = 0.142

EV of node D is the maximum.

Therefore best decision strategy is:  

C) Take the test, Bar him if the result is positive, do not bar him if the result is negative

EV of this decision strategy = 0.142

ANSWER: -11.1