Answer :
Sure! Let's proceed step by step.
First, let's complete the table:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & & \multicolumn{6}{|c|}{ Die } \\
\hline & & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \multirow{2}{*}{ Coin } & heads & [tex]$H-1$[/tex] & [tex]$H-2$[/tex] & [tex]$H-3$[/tex] & [tex]$H-4$[/tex] & [tex]$H-5$[/tex] & [tex]$H-6$[/tex] \\
\hline & tails & [tex]$T-1$[/tex] & [tex]$T-2$[/tex] & [tex]$T-3$[/tex] & [tex]$T-4$[/tex] & [tex]$T-5$[/tex] & [tex]$T-6$[/tex] \\
\hline
\end{tabular}
The table now shows all possible outcomes when Ela rolls a die and then flips a coin. Each cell represents a unique outcome in the sample space.
Next, let's determine the size of the sample space.
Since there are 6 possible outcomes for the die (1 through 6) and 2 possible outcomes for the coin (heads and tails), the size of the sample space is
[tex]\[ 6 \text{ die outcomes} \times 2 \text{ coin outcomes} = 12 \text{ possible outcomes}. \][/tex]
So, the size of the sample space is [tex]\(\boxed{12}\)[/tex].
First, let's complete the table:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & & \multicolumn{6}{|c|}{ Die } \\
\hline & & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \multirow{2}{*}{ Coin } & heads & [tex]$H-1$[/tex] & [tex]$H-2$[/tex] & [tex]$H-3$[/tex] & [tex]$H-4$[/tex] & [tex]$H-5$[/tex] & [tex]$H-6$[/tex] \\
\hline & tails & [tex]$T-1$[/tex] & [tex]$T-2$[/tex] & [tex]$T-3$[/tex] & [tex]$T-4$[/tex] & [tex]$T-5$[/tex] & [tex]$T-6$[/tex] \\
\hline
\end{tabular}
The table now shows all possible outcomes when Ela rolls a die and then flips a coin. Each cell represents a unique outcome in the sample space.
Next, let's determine the size of the sample space.
Since there are 6 possible outcomes for the die (1 through 6) and 2 possible outcomes for the coin (heads and tails), the size of the sample space is
[tex]\[ 6 \text{ die outcomes} \times 2 \text{ coin outcomes} = 12 \text{ possible outcomes}. \][/tex]
So, the size of the sample space is [tex]\(\boxed{12}\)[/tex].