Chapter 5.4, Problem 30P

a.

To determine

Define a formula for $P\left(n\right)$ in the given context of an application.

Answer to Problem 30P

The formula for the probability that the fourth successful crop occurs is $P\left(n\right)=C_{n-1,3}{\left(0.65\right)}^4{\left(0.35\right)}^{n-4}$.

Explanation of Solution

Calculation:

Let n follows a negative binomial distribution that represents the number of years in which the fourth successful crop occurswith success described as “crops help to repay loan” and failure described as “crops do not help to repay loan”.

The probability of success in a year is $p=0.65$.

The probability of failure is $q=0.35$.

The number of successful crops is $k=4$.

Negative binomial probability:

The probability that k^th successoccurs on n^th trial is as given below:

$P\left(n\right)=C_{n-1,k-1}{p}^k{q}^{n-k}$

Here,n is the number of trial in which k^th success occurs, k is the number of successes, p is the probability of success, and q is the probability of failure.

The formula for the probability that the fourth successful crop occursisas given below:

$\begin{array}{l}P\left(n\right)=C_{n-1,k-1}{p}^k{q}^{n-k}\\ =C_{n-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{n-4}\\ =C_{n-1,3}{\left(0.65\right)}^4{\left(0.35\right)}^{n-4}\end{array}$

Thus, the formula for the probability that the fourth successful crop occurs is $P\left(n\right)=C_{n-1,3}{\left(0.65\right)}^4{\left(0.35\right)}^{n-4}$.

b.

To determine

Calculate the given probabilities.

Answer to Problem 30P

The probability that the fourth successful crop occurs in 4 years is 0.1785.

The probability that the fourth successful crop occurs in 5 years is 0.2499.

The probability that the fourth successful crop occurs in 6 years is 0.2187.

The probability that the fourth successful crop occurs in 7 years is 0.1531.

Explanation of Solution

Calculation:

The probability that the fourth successful crop occurs in 4 years is as given below:

$\begin{array}{l}P\left(4\right)=C_{4-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{4-4}\\ =C_{3,3}{\left(0.65\right)}^4{\left(0.35\right)}^0\\ =\left(\frac{3!}{3!0!}\right){\left(0.65\right)}^4{\left(0.35\right)}^0\\ =0.1785\end{array}$

Thus, the probability that the fourth successful crop occurs in 4 years is0.1785.

The probability that the fourth successful crop occurs in 5 years is as given below:

$\begin{array}{l}P\left(5\right)=C_{5-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{5-4}\\ =C_{4,3}{\left(0.65\right)}^4{\left(0.35\right)}^1\\ =\left(\frac{4!}{3!1!}\right){\left(0.65\right)}^4{\left(0.35\right)}^1\\ =0.2499\end{array}$

Thus, the probability that the fourth successful crop occurs in 5 years is0.2499.

The probability that the fourth successful crop occurs in 6 years is as given below:

$\begin{array}{l}P\left(6\right)=C_{6-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{6-4}\\ =C_{6,3}{\left(0.65\right)}^4{\left(0.35\right)}^2\\ =\left(\frac{6!}{3!2!}\right){\left(0.65\right)}^4{\left(0.35\right)}^2\\ =0.2187\end{array}$

Thus, the probability that the fourth successful crop occurs in 6 years is0.2187.

The probability that the fourth successful crop occurs in 7 years is as given below:

$\begin{array}{l}P\left(7\right)=C_{7-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{7-4}\\ =C_{6,3}{\left(0.65\right)}^4{\left(0.35\right)}^3\\ =\left(\frac{6!}{3!3!}\right){\left(0.65\right)}^4{\left(0.35\right)}^3\\ =0.1531\end{array}$

Thus, the probability that the fourth successful crop occurs in 7 years is0.1531.

c.

To determine

Calculate the probability that a loan can be repaid by farm W within 4 to 7 years.

Answer to Problem 30P

The probability that the loan can be repaid by farm W within 4 to 7 years is 0.8002.

Explanation of Solution

Calculation:

The probability that the loan can be repaid by farm W within 4 to 7 years is calculated as given below:

$\begin{array}{l}P\left(4\le n\le 7\right)=P\left(4\right)+P\left(5\right)+P\left(6\right)+P\left(7\right)\\ =C_{4-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{4-4}+C_{5-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{5-4}+C_{6-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{6-4}\\ +C_{7-1,4-1}{\left(0.65\right)}^4{\left(0.35\right)}^{7-4}\\ =0.1785+0.2499+0.2187+0.1531\\ =\text{0}\text{.8002}\end{array}$

Thus, the probability that the loan can be repaid by farm W within 4 to 7 years is0.8002.

d.

To determine

Calculate the probability to farm for at least 8 years by farm W before they can repay the loan.

Answer to Problem 30P

The probability to farm for at least 8 years by farm W before they can repay the loan is 0.1998.

Explanation of Solution

Calculation:

The probability to farm for at least 8 years by farm W before they can repay the loan is calculated as given below:

$\begin{array}{l}P\left(n\ge 8\right)=1-P\left(n\le 7\right)\\ =1-P\left(4\le n\le 7\right)\\ =1-\text{0}\text{.8002}\\ =\text{0}\text{.1998}\end{array}$

Thus, the probability to farm for at least 8 years by farm W before they can repay the loan is0.1998.

e.

To determine

Calculate the expected value of n.

Calculate the standard deviationof n.

Answer to Problem 30P

The expected value of n is 6.15.

The standard deviation of n is 1.82.

Explanation of Solution

Calculation:

The expected value of n is calculated as given below:

$\begin{array}{l}\mu =\frac{k}{p}\\ =\frac{4}{0.65}\\ =6.15\end{array}$

Thus, the expected value of n is6.15.

The standard deviation of n is calculated as given below:

$\begin{array}{l}\sigma =\frac{\sqrt{kq}}{p}\\ =\frac{\sqrt{4\times 0.35}}{0.65}\\ =1.82\end{array}$

Thus, the standard deviationof n is 1.82.

Interpretation:

The random variable that represents the expected year in which the fourth successful crop occurs is 6.15 with a standard deviation of 1.82.