\begin{tabular}{|c|c|}
\hline
0 & 12 \\
\hline
1 & 19 \\
\hline
2 & 22 \\
\hline
3 & 25 \\
\hline
4 & 15 \\
\hline
5 or more & 7 \\
\hline
Total & 100 \\
\hline
\end{tabular}

a) The probability that the randomly selected home has 2 devices = [tex]$\square$[/tex]

b) The probability that the randomly selected home does not have 2 devices = [tex]$\square$[/tex]

c) The probability that the randomly selected home has more than 2 devices = [tex]$\square$[/tex]

d) The probability that the randomly selected home has 2 or fewer devices = [tex]$\square$[/tex]



Answer :

Let's solve each part of the question step-by-step:

Given the distribution of the number of devices per home:

[tex]\[ \begin{array}{|c|c|} \hline 0 & 12 \\ \hline 1 & 19 \\ \hline 2 & 22 \\ \hline 3 & 25 \\ \hline 4 & 15 \\ \hline 5 & 7 \\ \hline \text{Total} & 100 \\ \hline \end{array} \][/tex]

a) The probability that a randomly selected home has 2 devices:

The probability can be found by dividing the number of homes with 2 devices by the total number of homes.

[tex]\[ \text{Probability (2 devices)} = \frac{\text{Number of homes with 2 devices}}{\text{Total number of homes}} = \frac{22}{100} = 0.22 \][/tex]

So,

[tex]\[ \boxed{0.22} \][/tex]

b) The probability that a randomly selected home does not have 2 devices:

This probability is the complement of the probability of having 2 devices, which is 1 minus the probability of having 2 devices.

[tex]\[ \text{Probability (not 2 devices)} = 1 - \text{Probability (2 devices)} = 1 - 0.22 = 0.78 \][/tex]

So,

[tex]\[ \boxed{0.78} \][/tex]

c) The probability that a randomly selected home has more than 2 devices:

To find this probability, we sum the probabilities of having 3, 4, and 5 devices and then divide by the total number of homes.

[tex]\[ \text{Number of homes with more than 2 devices} = 25 + 15 + 7 = 47 \][/tex]

[tex]\[ \text{Probability (more than 2 devices)} = \frac{47}{100} = 0.47 \][/tex]

So,

[tex]\[ \boxed{0.47} \][/tex]

d) The probability that a randomly selected home has 2 or fewer devices:

To find this probability, we sum the probabilities of having 0, 1, and 2 devices and then divide by the total number of homes.

[tex]\[ \text{Number of homes with 2 or fewer devices} = 12 + 19 + 22 = 53 \][/tex]

[tex]\[ \text{Probability (2 or fewer devices)} = \frac{53}{100} = 0.53 \][/tex]

So,

[tex]\[ \boxed{0.53} \][/tex]

To summarize:
- a) The probability that the randomly selected home has 2 devices = [tex]\(\boxed{0.22}\)[/tex]
- b) The probability that the randomly selected home does not have 2 devices = [tex]\(\boxed{0.78}\)[/tex]
- c) The probability that the randomly selected home has more than 2 devices = [tex]\(\boxed{0.47}\)[/tex]
- d) The probability that the randomly selected home has 2 or fewer devices = [tex]\(\boxed{0.53}\)[/tex]