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eBook An oll company purchased an option on land in Alaska. Preliminary geologic

ID: 3073872 • Letter: E

Question

eBook An oll company purchased an option on land in Alaska. Preliminary geologic studies assigned the following prior probabilities. P(high-quality oil) - 0.4 P(medium-quality oil) 0.5 P(no oil) 0.1 a. What is the probability of finding oil (to 1 decimal)? b. After 200 feet of drilling on the first well, a soll test is taken. The probabilities of finding the particular type of soil identified by the test are given below. P(soiljhigh-quality oil) 0.4 P(soil medium-quality oil)-0.5 P(soil no oil) -0.1 Given the soil found in the test, use Bayes' theorem to compute the following revised probabilities (to 4 decimals). P(high-quality oil soil) P(medium-quality oil soil) P(no oil soil) What is the new probability of finding oil (to 4 decimals)?

Explanation / Answer

(a)

The probability of finding oil is:

P(oil) = P(high-quality oil) + P(medium-quality oil) = 0.4 + 0.5 = 0.9

(b)

P(high-quality oil| soil) = [P(soil | high-quality oil) P(high-quality oil) ] / [P(soil | high-quality oil) P(high-quality oil) +P(soil | medium-quality oil) P(medium-quality oil) +P(soil | no oil) P(no oil) ]

= [ 0.4 * 0.4 ] / [ 0.4*0.4 + 0.5 * 0.5 + 0.1 * 0.1 ] = 0.16 / [0.16 + 0.25 +0.01] = 0.16 /0.42 = 0.3810

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P(medium-quality oil| soil) = [P(soil | medium-quality oil) P(medium-quality oil) ] / [P(soil | high-quality oil) P(high-quality oil) +P(soil | medium-quality oil) P(medium-quality oil) +P(soil | no oil) P(no oil) ]

= [ 0.5 * 0.5 ] / [ 0.4*0.4 + 0.5 * 0.5 + 0.1 * 0.1 ] = 0.25 / [0.16 + 0.25 +0.01] = 0.25 /0.42 = 0.5952

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P(no oil| soil) = [P(soil | no oil) P(no oil) ] / [P(soil | high-quality oil) P(high-quality oil) +P(soil | medium-quality oil) P(medium-quality oil) +P(soil | no oil) P(no oil) ]

= [ 0.1 * 0.1 ] / [ 0.4*0.4 + 0.5 * 0.5 + 0.1 * 0.1 ] = 0.01 / [0.16 + 0.25 +0.01] = 0.01 /0.42 = 0.0238

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The new peobability of finding oil is:

P(high-quality oil| soil) + P(medium-quality oil| soil) = 0.3810 + 0.5952 = 0.9762

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