Patty is trying to determine which of two college courses to take. If she takes

Question

Patty is trying to determine which of two college courses to take. If she takes the Operations Research course, she believes that she has a 10% chance of receiving an A, a 40% chance for a B, and a 50% chance for a C. If Patty takes a statistics course, she has a 70% chance for a B, a 25% chance for a C, and a 5% chance for a D. Patty wants to explore her utility function for grades, and she sets U(A) = 4 and U(D) -1. Patty is indifferent between the following lotteries for grades. 0.25 1 C and L 0.75 She is also indifferent between 0.4 1 B and L 0.6 (A) If Patty wants to take the course that maximizes the expected utility of her final grade, which course should she take? (B) Patty believes that if she spends more time than regular on the statistics course, she might have a chance to get an A. Assuming she still has a 25% chance for a C, and 5% chance for a D, find the break- even probability she gets an A for the statistics course that makes her indifferent between the Operations Research course and the statistics course.

Explanation / Answer

First we need to find the utility function of Patty for the grades

We know Patty's utility for getting A and D as

U(A) =4

U(D)=1

We need to find Patty's utility for grades B and C

We know that Patty is indifferent between getting a C for certain and a lottery with 0.25 probability of getting an A and 0.75 probability of getting a D. We write the utility equations as

U(C) = 0.25*U(A) + 0.75*U(D) = 0.25*4 + 0.75*1 = 1.75

We also know that Patty is indifferent between getting a B for certain and a lottery with 0.40 probability of getting an A and 0.60 probability of getting a D. We write the utility equations as

U(C) = 0.4*U(A) + 0.6*U(D) = 0.4*4 + 0.6*1 = 2.2

So we have Patty's utility for grades A,B,C and D

U(A) =4

U(B) = 2.2

U(C)=1.75

U(D)=1

Now we calculate the expected utility of final grade for the course Operation Research. We know the following probabilities of grades for Operation research (OR)

Probability of getting an A, P(A) = 0.1

Probability of getting a B, P(B) = 0.40

Probability of getting a C, P(C) = 0.50

Patty has a zero probability of getting a D in OP

The expected utility for taking OR is

EU(OP) = P(A)*U(A) + P(B)*U(B) + P(C)*U(C) + P(D)*U(D)

EU(OR) = 0.10*4 + 0.40*2.2 + 0.50*1.75 + 0*1 = 2.155

Similarly we have the probability of Patty getting different grades in Statistics (STAT) as

Probability of getting an A, P(A) = 0

Probability of getting a B, P(B) = 0.70

Probability of getting a C, P(C) = 0.25

Probability of getting a D P(D) = 0.05

The expected utility for taking Statistics is

EU(STAT) = P(A)*U(A) + P(B)*U(B) + P(C)*U(C) + P(D)*U(D)

EU(OP) = 0*4 + 0.70*2.2 + 0.25*1.75 + 0.05*1 = 2.0275

We can see that Patty's expected utility for taking Operation Research is higher than for Statistics. Hence she should take Operation Research to maximize her utility of the final grade

b)

Let the probability of Patty getting an A in statistics is p and we know that her probability of getting a C, P(C) = 0.25 and getting a D, P(D) = 0.05. We know that

P(A)+P(B)+P(C)+P(D) =1 or

p+P(B)+0.25+0.05 =1

P(B) = 0.7-p

For Patty to be indifferent between Operation research and statistics, the expected utility of statitics needs to be equal to the expected utility of OR. Or

EU(STAT) = 2.155

We also know

EU(STAT) = P(A)*U(A) + P(B)*U(B) + P(C)*U(C) + P(D)*U(D), substituting probabilities and utilities we get

2.155 = p*4 + (0.7-p)*2.2 + 0.25*1.75 + 0.05*1

2.155 = 4p- 2.2p + 0.7*2.2 + 0.25*1.75 +0.05*1

1.8p = 2.155 - 2.0275

p = 0.071

That is the break even probability for getting an A in statistics, that will make get indifferent between OR and stat is 7.1%

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