Chapter 11.9, Problem 39PPSE

Solution

(a)

To calculate:

The probability distribution for ${(p+q)}^3$ and plot it.

The probability distribution for ${(p+q)}^3$ is shown below

X	P(X)
0	0.001
1	0.027
2	0.243
3	0.729

Given:

  $\begin{array}{l}p=0.9\\ q=0.1\end{array}$

Concept used:

The expansion of ${(p+q)}^3$ is shown below

  ${(p+q)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)p^0{q}^{n-0}+\left(\begin{array}{c}n\\ 1\end{array}\right)p^1{q}^{n-1}+\mathrm{..}+\left(\begin{array}{c}n\\ n\end{array}\right)p^n{q}^{n-n}$ where $\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

The expansion of ${(p+q)}^3$ is shown below

  $\begin{array}{c}{(}^p=\left(\begin{array}{c}3\\ 0\end{array}\right)p^0{q}^3+\left(\begin{array}{c}3\\ 1\end{array}\right)p^1{q}^2+\left(\begin{array}{c}3\\ 1\end{array}\right)p^2{q}^1+\left(\begin{array}{c}3\\ 3\end{array}\right)p^3{q}^0\\ =\frac{3!}{0!\left(3-0\right)!}{q}^3+\frac{3!}{1!\left(3-1\right)!}p{q}^2+\frac{3!}{2!\left(3-2\right)!}{p}^{2}q+\frac{3!}{3!\left(3-3\right)!}{p}^3\\ =q^3+3p{q}^2+3p^{2}q+p^3\end{array}$

  $\begin{array}{c}\text{For }p=0.9\text{ and }q=0.1\\ $ {(p+q)}^3={\left(0.1\right)}^3+3(0.9){(0.1)}^2+3{(0.9)}^2(0.1)+{(0.9)}^3$=0.001+0.027+0.243+0.729\end{array}$

Interpretation:

The probability distribution for $p=0.9\text{ and }q=0.1$ is shown below

X	P(X)
0	0.001
1	0.027
2	0.243
3	0.729

Enter the probability distribution in excel in two columns and insert a column chart to get the following

  

(b)

To calculate:

The probability distribution for ${(p+q)}^3$ and plot it.

The probability distribution for ${(p+q)}^3$ is shown below

X	P(X)
0	0.1664
1	0.4084
2	0.3341
3	0.0911

Given:

  $\begin{array}{l}p=0.45\\ q=0.55\end{array}$

Calculation:

The expansion of ${(p+q)}^3$ is shown below

  $\begin{array}{c}{(}^p=\left(\begin{array}{c}3\\ 0\end{array}\right)p^0{q}^3+\left(\begin{array}{c}3\\ 1\end{array}\right)p^1{q}^2+\left(\begin{array}{c}3\\ 1\end{array}\right)p^2{q}^1+\left(\begin{array}{c}3\\ 3\end{array}\right)p^3{q}^0\\ =\frac{3!}{0!\left(3-0\right)!}{q}^3+\frac{3!}{1!\left(3-1\right)!}p{q}^2+\frac{3!}{2!\left(3-2\right)!}{p}^{2}q+\frac{3!}{3!\left(3-3\right)!}{p}^3\\ =q^3+3p{q}^2+3p^{2}q+p^3\end{array}$

  $\begin{array}{c}\text{For }p=0.45\text{ and }q=0.55\\ $ {(p+q)}^3={\left(0.55\right)}^3+3(0.45){(0.55)}^2+3{(0.55)}^2(0.45)+{(0.45)}^3$=0.1664+0.4084+0.3341+0.0911\end{array}$

Interpretation:

The probability distribution for $p=0.45\text{ and }q=0.55$ is shown below

X	P(X)
0	0.1664
1	0.4084
2	0.3341
3	0.0911

Enter the probability distribution in excel in two columns and insert a column chart to get the following

  

(c)

To compare:

The graphs plotted in both part a and b.

The graphs plotted in both parts are different as the probability increases with number of success in part a where probability increases from 0 to 1 success and then decreased in plot in part b.

Interpretation:

With reference to both the plots, the probabilities for success 0, 1, 2, and 3 keep on increasing as the number of successes increases and the maximum probability is for the number of successes three when$p=0.9\text{ and }q=0.1$ . Similarly, when $p=0.45\text{ and }q=0.55$ , the probability of success increases from 0 to 1 and the probability of success decreases as the number of successes starts increasing from 1 to 3.