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Oil Example An oil company is considering a site for an exploratory well. If the rock strata underlying the site are characterized by what geologists call.

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Presentation on theme: "Oil Example An oil company is considering a site for an exploratory well. If the rock strata underlying the site are characterized by what geologists call."— Presentation transcript:

1 Oil Example An oil company is considering a site for an exploratory well. If the rock strata underlying the site are characterized by what geologists call a “dome” structure, the chances of finding oil are somewhat greater than if no dome structure exists. The probability of a dome structure is Pr(Dome)=0.6. The conditional probabilities of finding oil in this site are as follows. -100k Dome No Dome Dry (0.60) Low (0.25) High (0.15) Dry (0.2) Dry (0.85) Low (0.125) High (0.025) Low (0.8) 150k 500k -200k -100k 150k 500k 50k (0.908) (0.092) 52.50k k EMV=10k EMV=0 Site 1 Site 2 Pr(Dry|Dome) = 0.6 Pr(Low|Dome) = 0.25 Pr(High|Dome) = 0.15 Pr(Dry|No Dome) = 0.85 Pr(Low|No Dome) = Pr(High|No Dome) = 0.025

2 7.32 If the company could collect information from a drilling core sample and analyze it to determine whether a dome structure exists at site 1. A positive result would indicate the presence of dome, and a negative result would indicate the absence of a dome. The test is not perfect, however. Particularly, Pr(+|Dome)=0.99, and Pr(-|No Dome)=0.85. Use these probabilities and information given in the example, and Bayes’ theorem to find the posterior probabilities Pr(Dome|+) and Pr(Dome|-). If the test gives a positive result, which site should be selected? If the test result is negative, which site should be selected? Pr(Dome |+) = Pr(Dome |– ) =

3 52.50k k 52.50k k -100k Positive Dome Negative No Dome Dome No Dome Dry (0.60) Low (0.25) High (0.15) Dry (0.2) Dry (0.85) Low (0.125) High (0.025) Dry (0.60) Low (0.25) High (0.15) Dry (0.85) Low (0.125) High (0.025) Low (0.8) 150k 500k -200k -100k 150k 500k -100k 150k 500k -100k 150k 500k 50k (0.654) (0.346) (0.908) (0.092) (0.017) (0.983) Site 1 Site 2 EMV(Site 1|+) = EMV (Dome) P(Dome | +) + EMV (No dome) P(No dome | +) = (52.50 K) ( K) = $ K If the test gives a positive result, what is the expected money value of Site 1? EMV(Site 1|+)=? Because EMV(Site 1|+) > 0, site 1 should be selected EMV=0

4 52.50k k 52.50k k -100k Positive Dome Negative No Dome Dome No Dome Dry (0.60) Low (0.25) High (0.15) Dry (0.2) Dry (0.85) Low (0.125) High (0.025) Dry (0.60) Low (0.25) High (0.15) Dry (0.85) Low (0.125) High (0.025) Low (0.8) 150k 500k -200k -100k 150k 500k -100k 150k 500k -100k 150k 500k 50k (0.654) (0.346) (0.908) (0.092) (0.017) (0.983) Site 1 Site 2 EMV(Site 1|–)=? If the test gives a negative result, what is the expected money value of Site 1? EMV(Site 1|–) = EMV (Dome) P(Dome | –) + EMV (No dome) P(No dome | –) = (52.50 K) ( K) = -$ K Because EMV(Site 1| – ) < 0, site 2 should be selected

5 -100k Positive Dome Negative No Dome Dome No Dome Dry (0.60) Low (0.25) High (0.15) Dry (0.2) Dry (0.85) Low (0.125) High (0.025) Dry (0.60) Low (0.25) High (0.15) Dry (0.85) Low (0.125) High (0.025) Low (0.8) 150k 500k -200k -100k 150k 500k -100k 150k 500k -100k 150k 500k 50k (0.654) (0.346) (0.908) (0.092) (0.017) (0.983) Site 1 Site 2 Pr(Dry|Dome), Pr(Low|Dome), Pr(High|Dome), Pr(Dry|No Dome), Pr(Low|No Dome) and Pr(High|No Dome) are not affected by the test result. Thus, we can conclude that the test result and the amount oil in the well are conditionally independent given the information about the dome structure of the well.

6 7.33 Pr(+ and Dome)=? Pr(+ and Dome + Dry)=? Pr(Dome|+ and Dry)=? Pr(+ AND Dome) = Pr(+| Dome) Pr(Dome) = 0.99(0.60) = Pr(+ AND Dome and Dry) = Pr(Dry and + and Dome) = Pr(Dry | + AND Dome) Pr(+ AND Dome) Because the amount of oil and test result are conditionally independent given the dome structure, Pr(Dry | + AND Dome) =Pr(Dry| Dome) =0.6 Therefore, Pr(+ AND Dome AND Dry) = 0.60 (0.594) = Pr(Dome | + AND Dry) = Pr(+ AND Dry) = Pr(+ AND Dry AND Dome) + Pr(+ AND Dry AND No Dome) Pr(+ AND Dry AND No Dome) = Pr(Dry AND + and No Dome) = Pr(Dry | + AND No Dome) Pr(+ AND No Dome) = Pr(Dry | No Dome) Pr(+ and No Dome) Pr(+ AND No Dome) = Pr(+|No Dome) Pr(No Dome) =(1-0.85)(0.4) = 0.06 So Pr (+AND Dry AND No Dome) = 0.85(0.06) = Pr(+AND Dry) = =0.407 Pr(Dome | + AND Dry) = 0.356/0.407 = 0.875


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