To determine which sequence of coin flips is consistent with the theoretical model, let's break down the problem step by step.
First, we need to understand the theoretical model. The coin is biased such that it results in 2 heads in every 3 flips on average.
Given 12 flips, we can calculate the expected number of heads. We multiply [tex]\( \frac{2}{3} \)[/tex] by 12:
[tex]\[
\text{Expected number of heads} = \frac{2}{3} \times 12 = 8
\][/tex]
Thus, we need to identify the sequence that contains 8 heads out of 12 flips.
Now we will examine each sequence:
Sequence A:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Flip & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Result & T & H & T & H & T & T & T & H & T & T & H & H \\
\hline
\end{tabular}
\][/tex]
Number of heads (H) in sequence A: 5
Sequence B:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Flip & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Result & H & H & H & T & T & H & T & T & H & H & H & H \\
\hline
\end{tabular}
\][/tex]
Number of heads (H) in sequence B: 8
Sequence C:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Flip & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Result & H & T & T & T & H & T & T & T & T & H & T & T \\
\hline
\end{tabular}
\][/tex]
Number of heads (H) in sequence C: 3
Sequence D:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Flip & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Result & H & H & T & T & T & T & H & T & H & T & T & T \\
\hline
\end{tabular}
\][/tex]
Number of heads (H) in sequence D: 5
Among the sequences provided, the one which contains exactly 8 heads is Sequence B.
Thus, the correct answer is:
[tex]\[
\boxed{B}
\][/tex]
