Question

A petroleum company estimates that a reservoir holds oil with a prior probability of 60\%. It then acquires petrophysical data that suggests the presence of oil. If the petrophysical analysis is accurate with a probability of 70\%, the posterior probability of the presence of oil in \% is ________ (rounded off to two decimal places).

Hint:

Always use Bayes' theorem when updating prior probabilities with new evidence. Here, the test accuracy directly affects the posterior probability calculation.

Answer 1

Step 1: Define the events
Let: - \(O\): Oil is present in the reservoir.
- \(D\): Petrophysical data suggests presence of oil. We are given: \[ P(O) = 0.6 \quad \text{(prior probability of oil)} \] \[ P(\overline{O}) = 1 - 0.6 = 0.4 \] \[ P(D|O) = 0.7 \quad \text{(probability test shows oil when oil is present)} \] \[ P(D|\overline{O}) = 1 - 0.7 = 0.3 \quad \text{(test wrongly suggests oil when oil is absent)} \] Step 2: Apply Bayes' theorem
The posterior probability is given by: \[ P(O|D) = \frac{P(D|O) \cdot P(O)}{P(D|O) \cdot P(O) + P(D|\overline{O}) \cdot P(\overline{O})} \] Step 3: Substitute values
\[ P(O|D) = \frac{(0.7)(0.6)}{(0.7)(0.6) + (0.3)(0.4)} \] \[ P(O|D) = \frac{0.42}{0.42 + 0.12} \] \[ P(O|D) = \frac{0.42}{0.54} \] Step 4: Simplify the fraction
\[ P(O|D) = 0.7777\ldots \approx 0.7778 \] \[ P(O|D) = 77.78\% \] Step 5: Final Answer
Thus, the posterior probability of oil given the petrophysical data is: \[ \boxed{77.78\%} \]