91Ó°ÊÓ

In a binomial distribution, the probability of achieving success is determined. With the respective values of n (number of trials), p(number of successes), and q(number of failures), the value of $P\left(X\right)$is calculated with the formula $\text{p}\left(x\right)=\frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}$

When n = 4, x = 2, and p = .2, the value ofis 0.8, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.2\\ & =& 0.8\\ & & \end{array}$

The value of p(x),when the values of x, n, and p are 2, 4, and 0.2, respectively, is calculated below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(2\right)& =& \frac{4!}{2!\left(4-2\right)!}{\left(0.2\right)}^2{\left(0.8\right)}^{4-2}\\ & =& \frac{4×3×2×1}{2\left(2\right)!}{\left(0.2\right)}^2{\left(0.8\right)}^2\\ & =& 0.1536\\ & & \end{array}$

The calculated value of q , which is 0.8, has been used to find the value ofp(2)as 0.1536.

When n = 3, x = 0, and p = .7, the value of is 0, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.7\\ & =& 0.3\\ & & \end{array}$

The value of ,when the values of x, n, and pare0, 3, and 0.7, respectively, is calculated below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(0\right)& =& \frac{3!}{0!\left(3-0\right)!}{\left(0.7\right)}^0{\left(0.3\right)}^{3-0}\\ & =& \frac{3×2×1}{1\left(3\right)!}{\left(0.7\right)}^0{\left(0.3\right)}^3\\ & =& 0.027\\ & & \end{array}$

The calculated value of q, which is 0.3, has been used to find the value ofp(0)as 0.027.

c.

When n = 5, x = 3, and p = .1, the computed value of is 0.9, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.1\\ & =& 0.9\\ & & \end{array}$

Thevalue of ,when the values of x, n, and pare3, 5, and 0.1, respectively, is calculated below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(3\right)& =& \frac{5!}{3!\left(5-3\right)!}{\left(0.1\right)}^3{\left(0.9\right)}^{5-3}\\ & =& \frac{5×4×3×2×1}{6\left(2\right)!}{\left(0.1\right)}^3{\left(0.9\right)}^2\\ & =& 0.0081\\ & & \end{array}$

The computed value of q , which is 0.9, has been used to find the value of $p\left(3\right)$as 0.0081.

When n = 3, x = 1, and p = .9, the value ofis 0.1, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.9\\ & =& 0.1\\ & & \end{array}$

The value of $\text{p}\left(\text{x}\right)$,when the values of x, n, and pare1, 3, and 0.9, respectively, is calculated below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(1\right)& =& \frac{3!}{1!\left(3-1\right)!}{\left(0.9\right)}^1{\left(0.1\right)}^{3-1}\\ & =& \frac{3×2×1}{1\left(2\right)!}{\left(0.9\right)}^1{\left(0.1\right)}^2\\ & =& 0.027\\ & & \end{array}$

Therefore, the value of q , which is 0.1, has been used to find the value of $\text{p}\left(\text{1}\right)$as 0.027.

e.When n = 3, x = 1, and p = .3,the value of is 0.7, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.3\\ & =& 0.7\\ & & \end{array}$

The calculation of the value of p(x),when the values of x, n, and p are 1, 3, and 0.3, respectively, is shown below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(1\right)& =& \frac{3!}{1!\left(3-1\right)!}{\left(0.3\right)}^1{\left(0.7\right)}^{3-1}\\ & =& \frac{3×2×1}{1\left(2\right)!}{\left(0.3\right)}^1{\left(0.7\right)}^2\\ & =& 0.441\\ & & \end{array}$

The calculated value of q , which is 0.7, has been used to find the value ofp(1),which is 0.441.

When n = 4, x = 2, and p = .6, the computed value ofis 0.4, as shown below.

$\begin{array}{rcl}p+q& =& 1\\ q& =& 1-p\\ & =& 1-0.6\\ & =& 0.4\\ & & \end{array}$

The computed value of $p\left(x\right)$ ,when the values of x, n, and pare2, 4, and 0.6, respectively, is shown below.

$\begin{array}{rcl}p\left(x\right)& =& \frac{n!}{x!\left(n-x\right)!}{\left(p\right)}^x{\left(q\right)}^{n-x}\\ p\left(2\right)& =& \frac{4!}{2!\left(4-2\right)!}{\left(0.6\right)}^2{\left(0.4\right)}^{4-2}\\ & =& \frac{4×3×2×1}{2\left(2\right)!}{\left(0.6\right)}^2{\left(0.4\right)}^2\\ & =& 0.3456\\ & & \end{array}$

The computed value of q , which is 0.4, has been used to find the value ofp(2)as 0.3456.