Assume that hitting oil at one drilling location is independent of another, and

Question

Assume that hitting oil at one drilling location is independent of another, and that, in a particular region, the probability of success at an individual location is 0.2. What is the probability that a driller will hit oil on or before the third drilling? Is it likely that a driller will have to drill more than 5 times before hitting oil? (Give first, the probability of this event, and then answer yes, or no) Suppose the drilling company believes that a venture will be profitable if the number of wells drilled until the second success occurs is less than or equal to 7. Find the probability that the venture will be successful.

Explanation / Answer

Geometric Distribution

X - no. of drillings for first hit

P(X = x) = (0.2)*(0.8)^(x-1) x = 1,2,3,4,5,...

(a)

P(X<=3)

= P(x=1) + P(X=2) + P(X=3)

= (0.2) + (0.2)(0.8) + (0.2)(0.8)^2

=0.488

(b)

P(X>5)

= 1 - P(X<=5)

= 1 - (P(x=1) + P(X=2) + P(X=3) + P(x=4) + P(X=5) )

= 1 - ((0.2) + (0.2)(0.8) + (0.2)(0.8)^2 + (0.2)(0.8)^3 + (0.2)(0.8)^4)

= 1 -0.67232

=0.32768

Since the probability is considerable (32.8%) , YES it is likely

(c)

Now this is a Negative Binomial Distribution ( as no. of succes is greater than 1)

X - no. of wells drilled until 2nd success

P(X=x) = (x-1)C(1) *(0.2)^2*(0.8)^(x-2) x = 2,3,4,5,...

since nC1 = n

P(X=x) = (x-1)*(0.2)^2*(0.8)^(x-2) x = 2,3,4,5,...

P(X<=7)

= P(X=2) + P(X=3) + P(x=4) + P(X=5) + P(x=6) + P(X=7)

= (0.2)^2 + (2)*(0.2)^2*(0.8) + (3)*(0.2)^2*(0.8)^2 + (4)*(0.2)^2*(0.8)^3 + (5)*(0.2)^2*(0.8)^4 + (6)*(0.2)^2*(0.8)^5

=0.4232832

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