Question

Management of the Telemore Company is considering developing and marketing a new product. It is estimated to be twice as likely that the product would prove to be successful as unsuccessful. If it were successful, the expected profit would be $1,500,000. If unsuccessful, the expected loss would be $1,800,000. A marketing survey can be conducted at a cost of $300,000 to predict whether the product would be successful. Past experience with such surveys indicates that successful products have been predicted to be successful 80% of the time, whereas unsuccessful products have been predicted to be unsuccessful 70% of the time. a) (10 points) Find the unconditional probability that the research predicts the product to be successful. Also, find the unconditional probability that the research predicts the product to be unsuccessful. b) (10 points) Find the posterior probabilities of the respective states of nature for each of the two possible predictions from the market survey (your answer should have four probabilities). c) (40 points) Draw the decision tree, including labeling all the decision and outcome nodes and branches, payoffs on all branches, and probabilities on outcome branches. Add up payoffs along each path from the root to a leaf node to obtain the payoff for the leaf node. d) (40 points) Use the backward induction procedure to find the optimal policy to maximize expected payoff. To get full credit, record all the expected payoffs used during the process, use a double dash (||) to block rejected decisions and state the optimal policy in words.

Answer 1

Solution :

First, define the following states:

S = Actual successful
F = Actual unsuccessful
IS = Predicted successful
IF = Predicted unsuccessful

(a)

It is given that:

Priors: P(S) = 2/3 and P(F) = 1/3

Conditionals: P(IS | S) = 0.80; P(IF | F) = 0.70

Therefore, P(IS | F) = 1 - P(IF | F) = 1 - 0.7 = 0.3; P(IF | S) = 1 - P(IS | S) = 1 - 0.8 = 0.2

Joint probabilits:

P(S) * P(IS | S) = (2/3)*0.80 = 0.5333
P(F) * P(IS | F) = (1/3)*0.30 = 0.100

So, P(IS) = P(S) * P(IS | S) + P(F) * P(IS | F) = 0.5333 + 0.1 = 0.6333

Also,

P(S) * P(IF | S) = (2/3)*0.20 = 0.1333
P(F) * P(IF | F) = (1/3)*0.70 = 0.2333

P(IF) = P(S) * P(IF | S) + P(F) * P(IF | F) = 0.1333 + 0.2333 = 0.3667

So, the unconditional probabilites are: P(IS) = 0.6333; P(IF) = 0.3667

(b)

Posteriors = Corresponding Joints / Unconditionals:

P(S | IS) = P(S) * P(IS | S) / P(IS) = 0.5333 / 0.6333 = 0.842

P(F | IS) = P(F) * P(IS | F) / P(IS) = 0.100 / 0.6333 = 0.158

P(S | IF) = P(S) * P(IF | S) / P(IF) = 0.1333 / 0.3667 = 0.364

P(F | IF) = P(F) * P(IF | F) / P(IF) = 0.2333 / 0.3667 = 0.636

(c)

(d)

The optimal policy is to go for the product development without any survey with an expected payoff of 400 thousand.

The survey is not worth 300 thousand. It is only worth (320+300 - 400) = 220 thousand.

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