Question #157567

In a basin area where oil is likely to be found underneath the surface, there are 3 locations with 3 different types of earth compositions, say C1, C2, & C3. The probabilities for these 3 compositions are 0.5, 0.3, & 0.2 respectively. Further, it has been found from the past experience that after drilling of well at these locations, the probabilities of finding oil is 0.2, 0.4, & 0.3 respectively. Suppose, a well is drilled at a location, & it yields oil, what is the probability that the earth composition was C1?



1
Expert's answer
2021-01-28T05:24:46-0500

Let E1,E2,E3E_1,E_2,E_3 be three events such that , E1=E_1= The location with earth composition was C1C_1

E2=E_2= The location with earth composition was C2C_2

E3=E_3= The location with earth composition was C3C_3

Also let AA be an event such that , A=A= finding oil at the location

Then P(E1)=0.5,P(E2)=0.3,P(E3)=0.2P(E_1)=0.5,P(E_2)=0.3,P(E_3)=0.2 and P(A/E1)=0.2,P(A/E2)=0.4,P(A/E3)=0.3P(A/E_1)=0.2,P(A/E_2)=0.4,P(A/E_3)=0.3

Now we have to find P(E1/A).P(E_1/A).

According to Bayes' Theorem we know that , P(E1/A)=P(E1).P(A/E1)P(E1).P(A/E1)+P(E2).P(A/E2)+P(E3).P(A/E3)P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)}

\therefore The required probability is P(E1/A)=(0.5×0.2)(0.5×0.2)+(0.3×0.4)+(0.2×0.3)=1028=514P(E_1/A)=\frac{(0.5×0.2)}{(0.5×0.2)+(0.3×0.4)+(0.2×0.3)}=\frac{10}{28}=\frac{5}{14}


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