Answer to Question #195556 in Statistics and Probability for Mohammed Moro

Question

A geological study indicates that an exploration oil well should strike oil with probability 0.2.

(a). What is the probability that the third strike comes on the seventh well drilled?

(b). What assumptions did you make to obtain the answers to part (a)?

(c). Find the mean and variance of the number of wells that must be drilled if the company wants to set up three producing wells?

Answer 1

(a). Given that:

$p=0.2$

$\:q=1-p=1-0.2=0.8\:\:$

$r=3\:\:$

Negative binomial probability:

$p\left(y\right)=\begin{pmatrix}y-1\\ r-1\end{pmatrix}p^rq^{y-r}$

Evaluate:

$p\left(7\right)=\begin{pmatrix}7-1\\ \:3-1\end{pmatrix}0.2^30.8^{7-3}=0.0492$

The probability is $0.0492$ .

(b). The random variable Y is equal to the number of trail on which the $rth$ success occurs $(r=1,2,3,4,etc)$ .

The trials are independent and the probability of success is the same from trial to trial.

(c). We know that Y has a negative binomial probability density function with probability $0.2$ .

Then the probability mass function is:

$p\left(Y=y\right)=^{y-1}C_{r-1}p^rq^{y-r}$

Then, $E\left[Y\right]=\frac{r}{p}and$

$V\left(Y\right)=\frac{r\left(1-p\right)}{p^2}$

Now, $r=3$ and $\:p=0.2$ .

The mean of the negative binomial distribution is given by:

$E\left(Y\right)=\frac{r}{p}=\frac{3}{0.2}$

$E\left(Y\right)=15$

The variance of the negative binomial distribution is given by:

$V\left(Y\right)=\frac{r\left(1-p\right)}{p^2}$

$V\left(Y\right)=\frac{3\left(1-0.2\right)}{\left(0.2\right)^2}$

$V\left(Y\right)=\frac{3\left(0.8\right)}{0.04}$

$V\left(Y\right)=60$ .