\(\begin{array}{l}
\text{Gọi $X_n$ là tổng số chấm khi gieo đồng thời hai con xúc xắc}\\
\left(X_{n}= \overline{2,12}\right)\\
\text{Gọi $P_{n}$ là xác suất xuất hiện $X_{n}$}\\
+)\quad P_{2} = \dfrac{1}{36}\\
+)\quad P_{3} = \dfrac{1}{36} \times 2= \dfrac{1}{18}\\
+)\quad P_4 = \dfrac{1}{36}\times 3 = \dfrac{1}{12}\\
+)\quad P_5 = \dfrac{1}{36}\times 4 = \dfrac19\\
+)\quad P_6 = \dfrac{1}{36}\times 5 = \dfrac{5}{36}\\
+)\quad P_7 = \dfrac{1}{36}\times 6 = \dfrac16\\
+)\quad P_8 = \dfrac{1}{36}\times 5 = \dfrac{5}{36}\\
+)\quad P_9 = \dfrac{1}{36}\times 4 = \dfrac{1}{9}\\
+)\quad P_{10} = \dfrac{1}{36}\times 3 = \dfrac{1}{12}\\
+)\quad P_{11} = \dfrac{1}{36}\times 2 = \dfrac{1}{18}\\
+)\quad P_{12} =\dfrac{1}{36}\\
\text{Ta được bảng phân phối xác suất của biến $X$ như sau:}\\
\begin{array}{|c|c|c|}
\hline
X&2&3&4&5&6&7&8&9&10&11&12\\
\hline
P&\dfrac{1}{36}&\dfrac{1}{18}&\dfrac{1}{12}&\dfrac{1}{9}&\dfrac{5}{36}&\dfrac{1}{6}&\dfrac{5}{36}&\dfrac{1}{9}&\dfrac{1}{12}&\dfrac{1}{18}&\dfrac{1}{36}\\
\hline
\end{array}
\end{array}\)
