Chapter 4.2, Problem 3E

a.

To determine

Compute the value of $P\left(X=2\right)$.

Answer to Problem 3E

The probability of $X=2$ is 0.3456.

Explanation of Solution

Given info:

The distribution of the random variable X is binomial with $n=4$ and $p=0.6$

Calculation:

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$P\left(X=x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n$

Where, p is the probability of success.

Substitute $n=4,x=2\text{and }p=0.6$

$\begin{array}{c}P\left(X=2\right)=\left(\begin{array}{l}4\\ 2\end{array}\right){\left(0.6\right)}^2{\left(1-0.6\right)}^{4-2}\\ =\frac{4!}{2!\left(4-2\right)!}{\left(0.6\right)}^2{\left(0.4\right)}^2\\ =\left(6\right)\left(0.36\right)\left(0.16\right)\\ =0.3456\end{array}$

Thus,  the value of $P\left(X=2\right)=0.3456$.

b.

To determine

Compute the value of $P\left(X>2\right)$.

Answer to Problem 3E

The probability of $X>2$ is 0.2031.

Explanation of Solution

Given info:

The distribution of the random variable X is binomial with $n=8$ and $p=0.2$

Calculation:

$P\left(X>2\right)=1-P\left(X\le 2\right)$

Substitute $n=8,x=2\text{and }p=0.2$

$\begin{array}{c}P\left(X>2\right)=1-P\left(X=0\right)-P\left(X=1\right)-P\left(X=2\right)\\ =\left(\begin{array}{c}1-\left(\begin{array}{l}8\\ 0\end{array}\right){\left(0.2\right)}^0{\left(1-0.2\right)}^{8-0}-\left(\begin{array}{l}8\\ 1\end{array}\right){\left(0.2\right)}^1{\left(1-0.2\right)}^{8-1}\\ +\left(\begin{array}{l}8\\ 2\end{array}\right){\left(0.2\right)}^2{\left(1-0.2\right)}^{8-2}\end{array}\right)\\ =\left(\begin{array}{c}1-\frac{8!}{0!\left(8-0\right)!}{\left(0.2\right)}^0{\left(0.8\right)}^8-\frac{8!}{1!\left(8-1\right)!}{\left(0.2\right)}^1{\left(0.8\right)}^7\\ +\frac{8!}{2!\left(8-2\right)!}{\left(0.2\right)}^2{\left(0.8\right)}^6\end{array}\right)\\ =1-\left(1\right)\left(1\right)\left(0.16777216\right)-\left(8\right)\left(0.2\right)\left(0.2097152\right)-\left(28\right)\left(0.04\right)\left(0.262144\right)\end{array}$

                $\begin{array}{c}=1-0.1678-0.3355-0.2936\\ =0.2031\end{array}$

Thus, the value of $P\left(X>2\right)=0.2031$.

c.

To determine

Compute the value of $P\left(X\le 2\right)$.

Answer to Problem 3E

The value of $P\left(X\le 2\right)=0.6826$.

Explanation of Solution

Given info:

The distribution of the random variable X is binomial with $n=5$ and $p=0.4$

Calculation:

$P\left(X\le 2\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)$

Substitute $n=5,x=0,1,2\text{and }p=0.4$

$\begin{array}{c}P\left(X\le 2\right)=\left(\begin{array}{c}\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.4\right)}^0{\left(1-0.4\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.4\right)}^1{\left(1-0.4\right)}^{5-1}\\ +\left(\begin{array}{l}5\\ 2\end{array}\right){\left(0.4\right)}^2{\left(1-0.4\right)}^{5-2}\end{array}\right)\\ =\left(\begin{array}{c}\frac{5!}{0!\left(5-0\right)!}{\left(0.4\right)}^0{\left(0.6\right)}^5+\frac{5!}{1!\left(5-1\right)!}{\left(0.4\right)}^1{\left(0.6\right)}^4\\ +\frac{5!}{2!\left(5-2\right)!}{\left(0.4\right)}^2{\left(0.6\right)}^3\end{array}\right)\\ =\left(\begin{array}{c}\left(1\right)\left(1\right)\left(0.07776\right)+\left(5\right)\left(0.4\right)\left(0.1296\right)\\ +\left(10\right)\left(0.16\right)\left(0.216\right)\end{array}\right)\\ =0.0778+0.2592+0.3456\end{array}$

$=0.6826$

Thus,  the value of $P\left(X\le 2\right)=0.6826$.

d.

To determine

Compute the value of $P\left(3\le X\le 5\right)$.

Answer to Problem 3E

The value of $P\left(3\le X\le 5\right)=0.8119$

Explanation of Solution

Given info:

The distribution of the random variable X is binomial with $n=6$ and $p=0.7$

Calculation:

$P\left(3\le X\le 5\right)=P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)$

Substitute $n=6,x=3,4,5\text{ and }p=0.7$

$\begin{array}{c}P\left(3\le X\le 5\right)=\left(\begin{array}{c}\left(\begin{array}{l}6\\ 3\end{array}\right){\left(0.7\right)}^3{\left(1-0.7\right)}^{6-3}+\left(\begin{array}{l}6\\ 4\end{array}\right){\left(0.7\right)}^4{\left(1-0.7\right)}^{6-4}\\ +\left(\begin{array}{l}6\\ 5\end{array}\right){\left(0.7\right)}^5{\left(1-0.7\right)}^{6-5}\end{array}\right)\\ =\left(\begin{array}{c}\frac{6!}{3!\left(6-3\right)!}{\left(0.7\right)}^3{\left(0.3\right)}^3+\frac{6!}{4!\left(6-4\right)!}{\left(0.7\right)}^4{\left(0.3\right)}^2\\ +\frac{6!}{5!\left(6-5\right)!}{\left(0.7\right)}^5{\left(0.3\right)}^1\end{array}\right)\\ =\left(20\right)\left(0.343\right)\left(0.027\right)+\left(15\right)\left(0.2401\right)\left(0.09\right)+\left(6\right)\left(0.16807\right)\left(0.3\right)\\ =0.18522+0.324135+0.302526\end{array}$

$=0.8119$

Thus,  the value of $P\left(3\le X\le 5\right)=0.8119$.