Chapter 6.2, Problem 47PS

Solution

a.

To determine: the probability of $P\left(X\right)$ for $X=0,1,2,\mathrm{.......}$

The probabilities for the value of $X$ are:

  $\begin{array}{l}P\left(0\right)\approx 0.099\\ P\left(1\right)\approx 0.271\\ P\left(2\right)\approx 0.319\\ P\left(3\right)\approx 0.208\\ P\left(4\right)\approx 0.081\\ P\left(5\right)\approx 0.019\\ P\left(6\right)\approx 0.0025\\ P\left(7\right)\approx 0.0001\end{array}$

Given information:

An average of 7 gopher holes appears on the farm shown each week.

  

Concept Used:

Probability:

It is the ratio of number of ways it can happen to the total number of outcomes:

  $P=\frac{\text{Number}\text{of}\text{favouable}\text{outcomes}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

Binomial Experiment Probability:

The probability of k successes in the n trials for a binomial experiment is given by $P\left(k\text{successes}\right)=C{}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

Let x represents the number of animal pits in the market plot.

The total area of the land where the animal pits might appear is

  $\begin{array}{c}A=0.8(0.3+0.3)\\ =0.48\end{array}$

The area of the marked plot is made of a rectangle with sides 0.3 and 0.8-0.5, and a right triangle sides 0.3 and 0.8-0.5.

Therefore, the area of the plot is calculated as the sum of the areas of the rectangle and the triangle:

  $\begin{array}{c}{A}_p=0.3\left(0.8-0.5\right)+\frac{0.3\left(0.8-0.5\right)}{2}\\ =0.135\end{array}$

The general probability of a pit appearing in the plot is determine by dividing the area of the plot with the total area of the land.

  $\begin{array}{c}p=\frac{0.135}{0.48}\\ =0.28125\end{array}$

Now, calculate the chance for each number of pits or for each value of x , using the formula for binomial probability distribution.

  $P\left(0\right)=C{}_0{(0.28125)}^0{\left(1-0.28125\right)}^{7-0}$

  $=0.0991$

  $P\left(1\right)=C{}_1{(0.28125)}^1{\left(1-0.28125\right)}^{7-1}$

  $=0.271$

  $P\left(2\right)=C{}_2{(0.28125)}^2{\left(1-0.28125\right)}^{7-2}$

  $=0.319$

  $P\left(3\right)=C{}_3{(0.28125)}^3{\left(1-0.28125\right)}^{7-3}$

  $=0.208$

  $P\left(4\right)=C{}_4{(0.28125)}^4{\left(1-0.28125\right)}^{7-4}$

  $=0.081$

  $P\left(5\right)=C{}_5{(0.28125)}^5{\left(1-0.28125\right)}^{7-5}$

  $=0.019$

  $P\left(6\right)=C{}_6{(0.28125)}^6{\left(1-0.28125\right)}^{7-6}$

  $=0.0025$

  $P\left(7\right)=C{}_7{(0.28125)}^7{\left(1-0.28125\right)}^{7-7}$

  $=0.0001$

b.

To make: a table for showing the probability distribution for X.

X	0	1	2	3	4	5	6	7
P(X)	0.099	0.271	0.319	0.208	0.081	0.019	0.0025	0.0001

Given information:

An average of 7 gopher holes appears on the farm shown each week.

  

Concept Used:

Probability:

It is the ratio of number of ways it can happen to the total number of outcomes:

  $P=\frac{\text{Number}\text{of}\text{favouable}\text{outcomes}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

Binomial Experiment Probability:

The probability of k successes in the n trials for a binomial experiment is given by $P\left(k\text{successes}\right)=C{}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

From the previous part, the probabilities for the value of $X$ are:

  $\begin{array}{l}P\left(0\right)\approx 0.099\\ P\left(1\right)\approx 0.271\\ P\left(2\right)\approx 0.319\\ P\left(3\right)\approx 0.208\\ P\left(4\right)\approx 0.081\\ P\left(5\right)\approx 0.019\\ P\left(6\right)\approx 0.0025\\ P\left(7\right)\approx 0.0001\end{array}$

X	0	1	2	3	4	5	6	7
P(X)	0.099	0.271	0.319	0.208	0.081	0.019	0.0025	0.0001

c.

To make: a histogram for showing the probability distribution for X.

  

Given information:

An average of 7 gopher holes appears on the farm shown each week.

  

Concept Used:

Probability:

It is the ratio of number of ways it can happen to the total number of outcomes:

  $P=\frac{\text{Number}\text{of}\text{favouable}\text{outcomes}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

Binomial Experiment Probability:

The probability of k successes in the n trials for a binomial experiment is given by $P\left(k\text{successes}\right)=C{}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

From the previous part, the probabilities for the value of $X$ are:

  $\begin{array}{l}P\left(0\right)\approx 0.099\\ P\left(1\right)\approx 0.271\\ P\left(2\right)\approx 0.319\\ P\left(3\right)\approx 0.208\\ P\left(4\right)\approx 0.081\\ P\left(5\right)\approx 0.019\\ P\left(6\right)\approx 0.0025\\ P\left(7\right)\approx 0.0001\end{array}$

From the previous part, the table for showing the probability distribution for X

X	0	1	2	3	4	5	6	7
P(X)	0.099	0.271	0.319	0.208	0.081	0.019	0.0025	0.0001

The histogram is shown below: