The probability that a wildcat well will be productive is 1/13. Assume that a group is drilling wells in various parts of the country so that the status of one well has no bearing on that of any other. Let X denote the number of wells drilled to obtain the first strikes.

a. Verify that X is geometric, and identify the value of the parameter p.

b. What is the exact expression for the density for X?

c. what is the exact expression for the moment generating function for X?

d. What are the numerical values of E[x], E[x2], \sigma 2, and \sigma ?

e. Find P[X>=2]

a) p = 1 / 13

b) f(x) = ( 12 / 13 ) ^(n-1) * 1 / 13

c) M(x) =  1/13 / ( 1 - (12/13)*e^t)  

d) E(X) = 13 ,  E(X^2) =  325 , Var (X) = 156 , S.d = 12.49

e)  P(X >= 2) = 12/13

Step-by-step explanation:

Given:

- The probability that a wildcat well is productive p = 1/13

Find:

- identify the value of the parameter p.

- What is the exact expression for the density for X?

- what is the exact expression for the moment generating function for X?

- What are the numerical values of E[x], E[x2], \sigma 2, and \sigma ?

- Find P[X>=2]

Solution:

- Declaring a random variable X is the number of wells drilled to obtain the first strikes.

                                     X ~ Geo ( 1 / 13 )

- The probability of success is independent from successive trials. Where X denotes the number of successive trials till there is a success. Hence, the parameter p = 1 / 13.

- The probability density function of the geometric distribution for number f trails till first success is given by:

                               f(x) = ( 1 - p ) ^(n-1) * p

                               f(x) = ( 12 / 13 )^(n-1) * 1 / 13

- The moment generating expression for a Geometric distribution is given by:

                              M(x) =  p / ( 1 - (1-p)*e^t)  

                              M(x) =  1/13 / ( 1 - (12/13)*e^t)  

- The expected value E(X) of a geometric function is given by:

                              E(X) = 1 / p

                              E(X) = 1 / (1 / 13)

                              E(X) = 13

Where,

                              Var(X) = ( 1 - p ) / p^2

                              Var(X) = ( 12/13 )*13^2

                              Var(X) = 156  

                               S.d = sqrt(156) = 12.49

We know,

                              Var(X) = E(X^2) - [ E(X) ]^2

                               E(X^2) =  Var(X) + [ E(X) ]^2

                               E(X^2) =  156 + 13^2

                              E(X^2) =  325

- The required probability of P(X >= 2 ) can be computed using f(x)

                              P(X >= 2 ) = 1 - f(1)

                             P(X >= 2 ) = 1 - ( 12 / 13 ) ^(1-1) * 1 / 13

                              P(X >= 2) = 1 - 1/13 = 12/13

he answer is x=1x3 because x=1 to x=3