Hi Wei,
Let event B={a field will produce oil} and event A_{i} = { ith drilled
well will produce oil}.
It is given that P(B)=0.25 and P(AB)=0.8. Hence P(not A_{i}B)
= 10.8=0.2 for any i=1,2,... and P(not B) =1P(B) =0.75. Moreover, it
is obvious that P(not A_{i} not B)=1 for any i=1,2,...
We made two assumptions to solve this problem. The first is that if
oil is present in the field then the field is capable of producing oil.
The second is that the probability that a particular well produces oil
is independent of whether any previous well has produced oil, that is
the events A_{i} are independent.
a)
We need to calculate
P(B not A_{1})= Bayes formula = P( not A_{1}B)*P(B) / P(not
A_{1}).
To calculate P(not A) we use the formula of total probabilities:
P(not
A_{1})
= P(not A_{1}B)*P(B) + P(not A_{1}not
B)*P(not B)
= (0.2)*(0.25) + (1)(0.75)
= 0.8.
Substituting all numbers we obtain:
P(B not A_{1})= (0.2)*(0.25)/0.8 = 0.0625=6.25 %
b)
We need to calculate
P(B not A_{1} and not A_{2})= Bayes formula = P( not A_{1} and
not A_{2} B)*P(B) /
P(not A_{1} and not A_{2}).
To calculate P( not A_{1} and not A_{2} B) we use independence:
P( not A_{1} and not A_{2} B)= P( not A_{1}
B)*P( not
A_{2}
B) =(0.2)*(0.2)=0.04.
To calculate P(not A_{1} and not A_{2}) we use the
formula of total probabilities:
P(not A_{1} and not A_{2})
= P(not A_{1} and
not A_{2}B)*P(B) + P(not
A_{1} and not A_{2} not B)*P(not B)
= (0.04)*(0.25)
+(1)*(0.75)=0.76.
Substituting all numbers we obtain:
P(B not A_{1} and not A_{2})= (0.04)*(0.25) /(0.76)=0.0132
= 1.32%
c)
Now consider three wells and
calculate P(B
not A_{1} and not A_{2} and not A_{3})
If this is less than 1% then three well will suffice. If not try four wells.
d)
Certainly, it will be 1 (common sense). Is that what
you get when you apply Bayes Formula?
Andrei and Penny
