Answer :
Sure, let's tackle the problem step by step, keeping the context and the given data in mind.
We have a table that describes the relationship between [tex]\(\varepsilon\)[/tex] and [tex]\(P(1)\)[/tex]:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline $\varepsilon$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline $P(1)$ & end & Ess & 87 & $e 24$ & e1e & 008 \\ \hline \end{tabular} \][/tex]
And the net value is given to be 14.
Part (a): Find P(1)
To address this, we will simply observe the table for the given values of [tex]\(\varepsilon\)[/tex] and [tex]\(P(1)\)[/tex]:
- For [tex]\(\varepsilon = 0\)[/tex], [tex]\(P(1) = \text{end}\)[/tex]
- For [tex]\(\varepsilon = 1\)[/tex], [tex]\(P(1) = \text{Ess}\)[/tex]
- For [tex]\(\varepsilon = 2\)[/tex], [tex]\(P(1) = 87\)[/tex]
- For [tex]\(\varepsilon = 3\)[/tex], [tex]\(P(1) = e \, 24\)[/tex]
- For [tex]\(\varepsilon = 4\)[/tex], [tex]\(P(1) = e1e\)[/tex]
- For [tex]\(\varepsilon = 5\)[/tex], [tex]\(P(1) = 008\)[/tex]
Thus, the values of [tex]\(P(1)\)[/tex] for [tex]\(\varepsilon\)[/tex] from 0 to 5 are:
[tex]\[ \text{end, Ess, 87, e 24, e1e, 008} \][/tex]
The second part of the problem mentions the net value which is specified as 14, but it seems we are focused primarily on determining [tex]\(P(1)\)[/tex] values as tabulated above.
Therefore, our detailed step-by-step solution provides the [tex]\(P(1)\)[/tex] values for each given [tex]\(\varepsilon\)[/tex].
To summarize:
(a) When [tex]\(\varepsilon = 0\)[/tex], [tex]\(P(1) = \text{end}\)[/tex] \\
When [tex]\(\varepsilon = 1\)[/tex], [tex]\(P(1) = \text{Ess}\)[/tex] \\
When [tex]\(\varepsilon = 2\)[/tex], [tex]\(P(1) = 87\)[/tex] \\
When [tex]\(\varepsilon = 3\)[/tex], [tex]\(P(1) = e \, 24\)[/tex] \\
When [tex]\(\varepsilon = 4\)[/tex], [tex]\(P(1) = e1e\)[/tex] \\
When [tex]\(\varepsilon = 5\)[/tex], [tex]\(P(1) = 008\)[/tex]
These are the answers for [tex]\(P(1)\)[/tex] based on the given [tex]\(\varepsilon\)[/tex] values.
We have a table that describes the relationship between [tex]\(\varepsilon\)[/tex] and [tex]\(P(1)\)[/tex]:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline $\varepsilon$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline $P(1)$ & end & Ess & 87 & $e 24$ & e1e & 008 \\ \hline \end{tabular} \][/tex]
And the net value is given to be 14.
Part (a): Find P(1)
To address this, we will simply observe the table for the given values of [tex]\(\varepsilon\)[/tex] and [tex]\(P(1)\)[/tex]:
- For [tex]\(\varepsilon = 0\)[/tex], [tex]\(P(1) = \text{end}\)[/tex]
- For [tex]\(\varepsilon = 1\)[/tex], [tex]\(P(1) = \text{Ess}\)[/tex]
- For [tex]\(\varepsilon = 2\)[/tex], [tex]\(P(1) = 87\)[/tex]
- For [tex]\(\varepsilon = 3\)[/tex], [tex]\(P(1) = e \, 24\)[/tex]
- For [tex]\(\varepsilon = 4\)[/tex], [tex]\(P(1) = e1e\)[/tex]
- For [tex]\(\varepsilon = 5\)[/tex], [tex]\(P(1) = 008\)[/tex]
Thus, the values of [tex]\(P(1)\)[/tex] for [tex]\(\varepsilon\)[/tex] from 0 to 5 are:
[tex]\[ \text{end, Ess, 87, e 24, e1e, 008} \][/tex]
The second part of the problem mentions the net value which is specified as 14, but it seems we are focused primarily on determining [tex]\(P(1)\)[/tex] values as tabulated above.
Therefore, our detailed step-by-step solution provides the [tex]\(P(1)\)[/tex] values for each given [tex]\(\varepsilon\)[/tex].
To summarize:
(a) When [tex]\(\varepsilon = 0\)[/tex], [tex]\(P(1) = \text{end}\)[/tex] \\
When [tex]\(\varepsilon = 1\)[/tex], [tex]\(P(1) = \text{Ess}\)[/tex] \\
When [tex]\(\varepsilon = 2\)[/tex], [tex]\(P(1) = 87\)[/tex] \\
When [tex]\(\varepsilon = 3\)[/tex], [tex]\(P(1) = e \, 24\)[/tex] \\
When [tex]\(\varepsilon = 4\)[/tex], [tex]\(P(1) = e1e\)[/tex] \\
When [tex]\(\varepsilon = 5\)[/tex], [tex]\(P(1) = 008\)[/tex]
These are the answers for [tex]\(P(1)\)[/tex] based on the given [tex]\(\varepsilon\)[/tex] values.
