Answered

A group of dental researchers are testing the effects of acidic drinks on dental crowns. They have five containers of crowns labeled V, W, X, Y, and Z. They will randomly select one of the containers to be the control for the experiment by drawing one of five well-mixed slips of paper with the same labels from a hat. Which of the following is the probability model for the control container?

\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$V$[/tex] & [tex]$W$[/tex] & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] \\
\hline
0.20 & 0.20 & 0.20 & 0.20 & 0.20 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$V$[/tex] & [tex]$W$[/tex] & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] \\
\hline
0.20 & 0.10 & 0.30 & 0.20 & 0.20 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$V$[/tex] & [tex]$W$[/tex] & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] \\
\hline
5 & 5 & 5 & -5 & 5 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|}
\hline
[tex]$W$[/tex] & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] \\
\hline
0.25 & 0.25 & 0.25 & 0.25 \\
\hline
\end{tabular}



Answer :

GinnyAnswer
To find the correct probability model for the control container in this experiment, we need to understand that the selection is done randomly and each container has an equal chance of being chosen. Here’s a step-by-step solution:

1. Understanding the Problem:
- There are five containers labeled V, W, X, Y, and Z.
- A slip of paper corresponding to each container is put into a hat.
- One slip is randomly drawn to select the control container.
- Since the selection is random and each container is equally likely to be chosen, each has the same probability.

2. Finding the Probability Model:
- Since there are 5 containers and the selection is equally likely, the probability for each container to be selected is:
[tex]\[ \frac{1}{5} = 0.20 \][/tex]

3. Validating the Options:
- The correct probability model must assign the same probability (0.20) to each container: V, W, X, Y, Z.
- Let’s analyze the given options:

Option 1:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline V & W & X & Y & Z \\ \hline 0.20 & 0.20 & 0.20 & 0.20 & 0.20 \\ \hline \end{tabular} \][/tex]
This option correctly assigns a probability of 0.20 to each container, which is consistent with our expectation.

Option 2:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline V & W & X & Y & Z \\ \hline 0.20 & 0.10 & 0.30 & 0.20 & 0.20 \\ \hline \end{tabular} \][/tex]
This option assigns different probabilities to the containers, which contradicts the requirement of equal likelihood.

Option 3:
[tex]\[ \begin{tabular}{|l|r|r|r|r|} \hline v & w & x & y & z \\ \hline 5 & 5 & 5 & -5 & 5 \\ \hline \end{tabular} \][/tex]
This option does not provide probabilities but rather numerical values that do not sum to 1 or make sense in the context of probability.

Option 4:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline W & X & Y & Z \\ \hline 0.25 & 0.25 & 0.25 & 0.25 \\ \hline \end{tabular} \][/tex]
This option neglects container V and assigns equal probability to the remaining containers. This is incorrect because it does not account for all containers and does not match our setup.

Therefore, the correct probability model is:

[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline V & W & X & Y & Z \\ \hline 0.20 & 0.20 & 0.20 & 0.20 & 0.20 \\ \hline \end{tabular} \][/tex]