Question

Inactive oil and gas structures. U.S. federal regulations require that operating companies clear all inactive offshore oil and gas structures within 1 year after production ceases. Researchers at the Louisiana State University Center for Energy Studies gathered data on both active and inactive oil and gas structures in the Gulf of Mexico (Oil & Gas Journal, Jan. 3, 2005). They discovered that the Gulf of Mexico has 2,175 active and 1,225 idle (inactive) structures. The following table breaks down these structures by type (caisson, well protector, or fixed platform). Consider the structure type and active status of one of these oil/gas structures.

Structure type	Caisson	Well protector	Fixed platform	Totals
Active	503	225	1447	2175
Inactive	598	177	450	1225

a. List the simple events for this experiment.

b. Assign reasonable probabilities to the simple events.

c. Find the probability that the structure is active.

d. Find the probability that the structure is a well protector.

e. Find the probability that the structure is an inactive caisson.

Short answer

1. Sample events are AC, AW, AF, IC, IW, IF.
2. The probabilities are 0.148, 0.066, 0.426, 0.176, 0.052 and 0.132.
3. The probability of active structure is 0.640.
4. The probability that the structure is a well protector is 0.118.
5. The probability that the structure is an inactive caisson is 0.176.

Step-by-step solution

Important formula

The formula for probability is $\mathbf{P}\mathbf{=}\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{out}\mathrm{comes}}$.

List the simple events for this experiment.

a)

The sample events are

A=active structure

I= inactive structure

C=caisson structure

W=well protector structure

F= fixed platform structure.

So, sample events are AC, AW, AF, IC, IW, IF.

Assign reasonable probabilities to the simple events.

b)

The probabilities are

$\begin{array}{c}P\left(AC\right)=\frac{503}{3400}\\ =0.148\end{array}$

$\begin{array}{c}P\left(AW\right)=\frac{225}{3400}\\ =0.066\end{array}$

$\begin{array}{c}P\left(AF\right)=\frac{1447}{3400}\\ =0.426\end{array}$

$\begin{array}{c}P\left(IC\right)=\frac{598}{3400}\\ =0.176\end{array}$

$\begin{array}{c}P\left(IW\right)=\frac{177}{3400}\\ =0.052\end{array}$

$\begin{array}{c}P\left(IF\right)=\frac{450}{3400}\\ =0.132\end{array}$

Thus, the probabilities are 0.148, 0.066, 0.426, 0.176, 0.052 and 0.132.

The probability that the structure is active.

c)

$\begin{array}{c}\mathrm{P}\left(\text{activestructure}\right)=\frac{503+225+1447}{2175}\\ =\frac{2175}{3400}\\ =0.640\end{array}$

Thus, the probability of active structure is 0.640.

The probability that the structure is a well protector.

d)

$\begin{array}{c}\mathrm{P}\left(\mathrm{W}\right)=\mathrm{P}(\mathrm{AW}+\mathrm{IW})\\ =0.066+0.052\\ =0.118\end{array}$

Hence, the probability that the structure is a well protector is 0.118.

The probability that the structure is an inactive caisson.

e)

$\begin{array}{c}\mathrm{P}\left(\text{inactivecaisson}\right)=\frac{598}{3400}\\ =0.176\end{array}$

Therefore, the probability that the structure is an inactive caisson is 0.176.