Answer to Question #150500 in Statistics and Probability for Unknown

Question

An oil company conducts a geological study that indicates that an exploratory oil well in a certain
region should have 2r% chance of striking oil where r is the last three digits of your registration
number. If eight wells are drilled
a) What is the probability that the oil strikes in the third well?
b) What is the probability that oil strikes in at least five of the eight wells?
c) What is the mean value of not striking oil in a well?

Answer 1

a) To find the requested probability, we need to find $P(X=3).$

$X$ is a geometric random variable, since we are only looking for one success.

$$P(X=x)=(1-p)^xp, x=0,1,2,...$$

Given $p=0.02r, r$ is the last three digits of your registration number

$$P(X=3)=(1-0.02r)^3(0.02r)$$

b) Let $X$ be the number of successful strikes if $n$ wells are dug: $X\sim Bin(n,p)$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Given $n=8, p=0.02r$

$$P(X\geq5)=P(X=5)+P(X=6)$$

$$+P(X=7)+P(X=8)$$

$$=\dbinom{8}{5}(0.02r)^5(1-0.02r)^{8-5}+\dbinom{8}{6}(0.02r)^6(1-0.02r)^{8-6}$$

$$+\dbinom{8}{7}(0.02r)^7(1-0.02r)^{8-7}+\dbinom{8}{8}(0.02r)^8(1-0.02r)^{8-8}$$

$$=56(0.02r)^5(1-0.02r)^3+28(0.02r)^6(1-0.02r)^2$$

$$+8(0.02r)^7(1-0.02r)+(0.02r)^8$$

3) Let $X$ be the number of nonsuccessful strikes if $n$ wells are dug: $X\sim Bin(n,p)$

Given $1-p=1-0.02r$ .Then he mean value of not striking oil in a well if $n$ wells are dug is

$$\mu=n(1-0.02r)$$