Chapter 5, Problem 5.3.15RE

Let x be a binomial random variable with n = 12 and p = 0.3. Find the following:

a. P(X = 8)

b. P(X˂5)

c. P(X ≥ 10)

d. P(4 ˂ X ≤ 9)

To determine

To calculate: The probability of X as 8 whenn is 12 and p is 0.3

Answer to Problem 5.3.15RE

The probability, $P\left(X=8\right)=\underset{\_}{0.008}$

Explanation of Solution

Given info:

The number of trials n is 12 andp is 0.3.

Calculation:

The binomial distribution formula is,

$P\left(X\right)=\frac{n!}{\left(n-X\right)!X!}{p}^X{q}^{n-X}$

Where, n is the number of trials, x is the number of successes among n trials, p is the probability of success and q is the probability of failure.

Substitute n as 12, p as 0.3 and q as $0.7\left(=1-0.3\right)$

$\begin{array}{c}P\left(8\right)=\frac{12!}{\left(12-8\right)!8!}{\left(0.3\right)}^8{\left(0.7\right)}^{12-8}\\ =495\left(0.000065\right)\left(0.2401\right)\\ \approx 0.008\end{array}$

To determine

To calculate: The probability of X less than 5 whenn is 12 and p is 0.3

Answer to Problem 5.3.15RE

The probability, $P\left(X<5\right)=\underset{\_}{0.724}$

Explanation of Solution

Calculation:

Substitute n as 12, X as 0, 1, 2, 3 and 4 p as 0.3 and q as $0.7\left(=1-0.3\right)$

$\begin{array}{c}P\left(X<5\right)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)\\ =\left[\begin{array}{l}\frac{12!}{\left(12-0\right)!0!}{\left(0.3\right)}^0{\left(0.7\right)}^{12-0}+\frac{12!}{\left(12-1\right)!1!}{\left(0.3\right)}^1{\left(0.7\right)}^{12-1}+\frac{12!}{\left(12-2\right)!2!}{\left(0.3\right)}^2{\left(0.7\right)}^{12-2}+\\ \frac{12!}{\left(12-3\right)!3!}{\left(0.3\right)}^3{\left(0.7\right)}^{12-3}+\frac{12!}{\left(12-4\right)!4!}{\left(0.3\right)}^4{\left(0.7\right)}^{12-4}\end{array}\right]\\ =\left[\begin{array}{l}1\left(1\right)\left(0.0138\right)+12\left(0.3\right)\left(0.0197\right)+66\left(0.09\right)\left(0.0282\right)+\\ 220\left(0.027\right)\left(0.0138\right)+495\left(0.0081\right)\left(0.0576\right)\end{array}\right]\\ =0.0138+0.0709+0.1675+0.24+0.2309\end{array}$

                   $=0.724$

To determine

To calculate: The probability of X greater than or equal to10whenn is 12 and p is 0.3

Answer to Problem 5.3.15RE

The probability, $P\left(X\ge 10\right)=\underset{\_}{0.0002}$

Explanation of Solution

Calculation:

Substitute n as 12, X as 10, 11 and12 p as 0.3 and q as $0.7\left(=1-0.3\right)$

$\begin{array}{c}P\left(X\ge 10\right)=P\left(10\right)+P\left(11\right)+P\left(12\right)\\ =\left[\begin{array}{l}\frac{12!}{\left(12-10\right)!10!}{\left(0.3\right)}^{10}{\left(0.7\right)}^{12-10}+\frac{12!}{\left(12-11\right)!11!}{\left(0.3\right)}^{11}{\left(0.7\right)}^{12-11}+\\ \frac{12!}{\left(12-12\right)!12!}{\left(0.3\right)}^{12}{\left(0.7\right)}^{12-12}\end{array}\right]\\ =66\left(5.9049\times {10}^{-6}\right)\left(0.49\right)+12\left(1.77147\times {10}^{-6}\right)\left(0.7\right)+1\left(5.31441\times {10}^{-7}\right)\left(1\right)\\ =0.0002+0+0\end{array}$

                    =0.0002

To determine

To calculate: The probability of X greater than 4 and less than or equal to 9 whenn is 12 and p is 0.3

Answer to Problem 5.3.15RE

The probability, $P\left(4<X\le 9\right)=\underset{\_}{0.276}$

Explanation of Solution

Calculation:

Substitute n as 12, X as 5, 6, 7, 8 and 9 p as 0.3 and q as $0.7\left(=1-0.3\right)$

$\begin{array}{c}P\left(4<X\le 9\right)=P\left(5\right)+P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)\\ =\left[\begin{array}{l}\frac{12!}{\left(12-5\right)!5!}{\left(0.3\right)}^5{\left(0.7\right)}^{12-5}+\frac{12!}{\left(12-6\right)!6!}{\left(0.3\right)}^6{\left(0.7\right)}^{12-6}\\ +\frac{12!}{\left(12-7\right)!7!}{\left(0.3\right)}^7{\left(0.7\right)}^{12-7}+\frac{12!}{\left(12-8\right)!8!}{\left(0.3\right)}^8{\left(0.7\right)}^{12-8}+\frac{12!}{\left(12-9\right)!9!}{\left(0.3\right)}^9{\left(0.7\right)}^{12-9}\end{array}\right]\\ =\left[\begin{array}{l}792\left(0.0024\right)\left(0.0824\right)+924\left(0.0007\right)\left(0.1176\right)+792\left(0.0002\right)\left(0.1681\right)+\\ 495\left(0.00007\right)\left(0.2401\right)+220\left(0.00002\right)\left(0.343\right)\end{array}\right]\\ =0.1585+0.0792+0.0291+0.0078+0.0015\end{array}$ $=0.276$