Ans: A
Note that the sample space is as follow:
Note that the sample space is as follow:
$\begin{array}{|c|c|c|c|c|c|c|} \hline
& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ \hline
2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ \hline
3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ \hline
4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ \hline
5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ \hline
6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \\ \hline
\end{array}$
Therefore, the probability of getting the sum of the two numbers thrown is $7$
$\begin{array}{cl}
= & \dfrac{6}{36} \\
= & \dfrac{1}{6}
\end{array}$
Hence, the expected gain of the game
$\begin{array}{cl}
= & \dfrac{1}{6} \times 36 + \left( 1 – \dfrac{1}{6} \right) \times 6 \\
= & \$ 11
\end{array}$
