Chapter 6, Problem 21E

FILE In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that:

1. a. All nine have large-screen TVs?
2. b. Less than five have large-screen TVs?
3. c. More than five have large-screen TVs?
4. d. At least seven homes have large-screen TVs?

To determine

Compute the probability that all nine have large-screen TVs.

Answer to Problem 21E

The probability that all nine have large-screen TVs is 0.3874.

Explanation of Solution

The formula to find the binomial probability is as follows:

$\begin{array}{l}P\left(X\right)=C{}_x{\pi }^x{\left(1-\pi \right)}^{\left(n-x\right)}\\ \text{where, }C\text{ is the combination}\text{.}\\ n\text{ is the number of trials}\text{.}\\ X\text{ is the random variable}\text{.}\\ \pi \text{ is the probability of success}\text{.}\end{array}$

Here, n=9; π=0.90.

The probability that all nine have large-screen TVs is calculated as follows:

$\begin{array}{c}P\left(9\right)=C{}_9{\left(0.90\right)}^9{\left(1-0.90\right)}^{\left(9-9\right)}\\ =\frac{9!}{9!\left(9-9\right)!}\times {0.90}^9\times {0.10}^0\\ =1\times 0.3874\times 1\\ =0.3874\end{array}$

Therefore, the probability that all nine have large-screen TVs is 0.3874.

To determine

Compute the probability that less than five have large-screen TVs.

Answer to Problem 21E

The probability that less than five have large-screen TVs is 0.0009.

Explanation of Solution

The probability that less than five have large-screen TVs is calculated as follows:

$\begin{array}{c}P\left(x<5\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)\\ =\left\{\begin{array}{c}\left[C{}_0{\left(0.90\right)}^0{\left(1-0.90\right)}^{\left(9-0\right)}\right]+\left[C{}_1{\left(0.90\right)}^1{\left(1-0.90\right)}^{\left(9-1\right)}\right]+\\ \left[C{}_2{\left(0.90\right)}^2{\left(1-0.90\right)}^{\left(9-2\right)}\right]+\left[C{}_3{\left(0.90\right)}^3{\left(1-0.90\right)}^{\left(9-3\right)}\right]+\\ \left[C{}_4{\left(0.90\right)}^4{\left(1-0.90\right)}^{\left(9-4\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{c}\left(\frac{9!}{0!\left(9-0\right)!}\times {0.90}^0\times {0.10}^9\right)+\left(\frac{9!}{1!\left(9-1\right)!}\times {0.90}^1\times {0.10}^8\right)+\\ \left(\frac{9!}{2!\left(9-2\right)!}\times {0.90}^2\times {0.10}^7\right)+\left(\frac{9!}{3!\left(9-3\right)!}\times {0.90}^3\times {0.10}^6\right)+\\ \left(\frac{9!}{4!\left(9-4\right)!}\times {0.90}^4\times {0.10}^5\right)\end{array}\right\}=1\times 0.3874\times 1\\ =0.0000+0.0000+0.0000+0.0001+0.0008\\ =0.0009\end{array}$

Therefore, the probability that less than five have large-screen TVs is 0.0009.

To determine

Calculate the probability that more than five have large-screen TVs.

Answer to Problem 21E

The probability that more than five have large-screen TVs is 0.0083.

Explanation of Solution

The probability that more than five have large-screen TVs is calculated as follows:

$\begin{array}{c}P\left(x>5\right)=1-\left[P\left(x=5\right)+P\left(x<5\right)\right]\\ =1-\left[\left(\frac{9!}{5!\left(9-5\right)!}\times {0.90}^5\times {0.10}^4\right)+0.0009\right]\left[\text{from Part (b), }P\left(x<5\right)=0.0009\right]\\ =1-0.0074+0.0009\\ =0.0083\end{array}$

Therefore, the probability that more than five have large-screen TVs is 0.0083.

To determine

Calculate the probability that at least seven homes have large-screen TVs.

Answer to Problem 21E

The probability that at least seven homes have large-screen TVs is 0.947.

Explanation of Solution

The probability that at least seven homes have large-screen TVs is calculated as follows:

$\begin{array}{c}P\left(x\ge 7\right)=P\left(x=7\right)+P\left(x=8\right)+P\left(x=9\right)\\ =\left\{\begin{array}{c}\left[C{}_7{\left(0.90\right)}^7{\left(1-0.90\right)}^{\left(9-7\right)}\right]+\left[C{}_8{\left(0.90\right)}^8{\left(1-0.90\right)}^{\left(9-8\right)}\right]+\\ \left[C{}_9{\left(0.90\right)}^9{\left(1-0.90\right)}^{\left(9-9\right)}\right]\end{array}\right\}\\ =\left[\begin{array}{l}\left(\frac{9!}{7!\left(9-7\right)!}\times {0.90}^7\times {0.10}^2\right)+\left(\frac{9!}{8!\left(9-8\right)!}\times {0.90}^8\times {0.10}^1\right)+\\ \left(\frac{9!}{9!\left(9-9\right)!}\times {0.90}^9\times {0.10}^0\right)\end{array}\right]\\ =0.1722+0.3874+0.3874\\ =0.947\end{array}$

Therefore, the probability that at least seven homes have large-screen TVs is 0.947.