Question

To save on expenses, Rona and Jerry agreed to form a carpool for traveling to and from work. Rona prefers to use the somewhat longer but more consistent Queen City Avenue. Although Jerry prefers the quicker expressway, he agreed with Rona that they should take Queen City Avenue if the expressway has a traffic jam. The following payoff table provides the one-way time estimate in minutes for traveling to or from work: State of Nature Expressway Open Expressway Jammed Decision Alternative Sj 52 Queen City Avenue, de 30 30 Expressway, d2 25 45 Based on their experience with traffic problems, Rona and Jerry agreed on a 0.35 probability that the expressway would be jammed. In addition, they agreed that weather seemed to affect the traffic conditions on the expressway. Let C-clear O overcast Rrain The following conditional probabilities apply: PC 15.) -0.8 POIS) - 0.2 PUR | ) - 0.0 PLC | 52) = 0.1 POIN2) 0.3 PIR I $2) - 0.6 (a) Use Bayes' theorem for probability revision to compute the probability of each weather condition and the conditional probability of the expressway being open, 51, or jammed, 5 2. given each weather condition. If required, round your answers to three decimal places. If your answer is zero enter "o. P(SIC): P(s): Pls: IO) Ps 10): Psi IR): e ALRH

Answer 1

Sol:

From the information, observe that Rona and Jerry agreed to form a carpool for travelling to and from work. Rona prefers to use the somewhat longer but more consistent Queen City Avenue. The payoff table is as follows: + State of Nature Express Way Open Expressway Jammed S S2 30 Decision Alternative Queen City Avenue, d. Expressway, d 30 25 45 Consider C, O and R are the variable that represent clear, overcast and rain respectively. Based on their previous experience with traffic problems, Rona and Jerry agreed on a 0.15 probability that the expressway would be jammed. That is, P(S2) = 0.15 P(s) = 1 - P(32) = 1-0.15 = 0.85 The conditional probabilities are as follows: P(CS) = 0.80 P(C|s2)=0.10 P(O S;)=0.20 P(01 S2)=0.30 P(RS) = 0.00 PR 5) = 0.60

a) Calculate P(SC) using Bayes Theorem. P(S)P(CIS) P(SC)= P(S)P(C|S;)+P(92)P(C|S2) 0.85 x 0.80 (0.85x0.80)+ +(0.15 x 0.10) 0.68 0.695 = 0.98 Calculate P(S2C) using Bayes Theorem. P(S)P(C|S) P(SC)= P(S)P(C|S1)+P(-2)P(C|S2) 0.15 0.10 (0.85x0.80)+(0.15x0.10) 0.015 0.695 = 0.02 Calculate P($10) using Bayes theorem. P(S)P(OS) P($10)= P(s)P(O| S1)+ P(32)P(0192) 0.85 x 0.20 (0.85 x 0.20)+(0.15)(0.30) 0.17 0.22 = 0.79

Calculate P(S210) using Bayes theorem. P($20 P(S2)P(0 s2) P(s)P(0|s)+P(32)P(0|s2) 0.15 x 0.30 (0.85x0.20)+ +(0.15)(0.30) 0.05 0.22 = 0.21 Calculate P(SR) using Bayes theorem. = P(S)P(RUS) P(SR) = P(s)P(R|S)+P(S2)P(R|S2) 0.85 x 0.0 (0.85 x 0.0)+(0.15x0.60) 0 0.09 = 0 =

Calculate P(SR) using Bayes theorem. + P(s)P(R|S2) P(S2R) = P(S)P(R| S)+ P(92)P(R|S2) 0.15 x 0.60 (0.85x0.0)+(0.15 x 0.60) 0.09 0.09 = 1 b) Calculate the probability for the individual events. P(C)= P(s)P(C|S)+P(S2)P(C|s2) = (0.85x0.80)+(0.15x0.10) = 0.68 +0.015 = 0.695 P(0)=P(s)P(01 sz) + P(s)P(01 s2) = (0.85x0.20)+(0.15)(0.30) = 0.17 +0.045 = 0.215 P(R) = P(S1)P(R| Sz)+P(92)P(R|S2) = (0.85x0.0)+(0.15x0.60) = 0.0 +0.09 = 0.09

Draw the tree diagram. 30 0.98 s1 s2 di 0.02 30 25.432 d2 25 0.98 s1 0.02 25.432 S2 0.695 с 45 30 0.791 si di 0.209 s2 26.6502 O 0.215 30 29.186 30 25 d2 0.791 29.186 0.209 s2 R 0.09 45 30 30 s1 di 1 30 30 -25 $10 45 d2 S2 1 45 The expected value of d, at Cis, C(d)=(0.98x30)+(0.02 x 30) = 29.4 +0.06 = 30

The expected value of d, at Cis, C(d)= (0.9825)+(0.02 x 45) = 24.5 +0.9 = 25.40 = Therefore, the expected value of C is taken as the minimum values of 30 and 25.40 That is, E(C)= Min(30,25.40) = 25.40 The expected value of d, at O is, 0(d) = (0.791x 0.30)+(0.209 x 30) = 23.73 + 6.27 = 30

The expected value of d, at O is, 0(dz) = (0.791x0.25)+(0.209 x 45) = 19.775 +9.405 = 29.180 Therefore, the expected value of O is taken as the minimum values of 30 and 29.180 That is, E(O)= Min(30,29.180) = 29.180 The expected value of d, at Ris, R(d)=(0x0.30)+(1x30) = 0+30 = 30 The expected value of dat R is, R(dz)=(0x 25) + (1x 45) = 0 +45 = 45 Therefore, the expected value of O is taken as the minimum values of 30 and 45 That is, E(R)= Min(30,45) = 45

c) From the tree diagram, observe that take the expressway unless there is rain, if rain, take Queen City Avenue. Hence, the correct option is (ii). The expected travel time is, Expected value = P(C)E(C)+ P(O)E(0)+ P(R)E(R) = (0.695x 25.432)+(0.215x 29.180)+(0.09x 30) = 17.66 +6.24 +2.7 = 26.650

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