Bài tập 28 trang 76 SGK Toán 11 NC

Bài tập 28 trang 76 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\Omega  = \{ \left( {1;1} \right),\left( {1;2} \right),\left( {1;3} \right),\left( {1;4} \right),\\
\left( {1;5} \right),\left( {1;6} \right),\left( {2;1} \right),\left( {2;2} \right),
\end{array}\\
{\left( {2;3} \right),\left( {2;4} \right),\left( {2;5} \right),\left( {2;6} \right),}\\
\begin{array}{l}
\left( {3;1} \right),\left( {3;2} \right),\left( {3;3} \right),\left( {3;4} \right),\\
\left( {3;5} \right),\left( {3;6} \right),\left( {4;1} \right),\left( {4;2} \right),
\end{array}\\
{\left( {4;3} \right),\left( {4;4} \right),\left( {4;5} \right),\left( {4;6} \right),}\\
\begin{array}{l}
\left( {5;1} \right),\left( {5;2} \right),\left( {5;3} \right),\left( {5;4} \right),\\
\left( {5;5} \right),\left( {5;6} \right),\left( {6;1} \right),\left( {6;2} \right),
\end{array}\\
{\left( {6;3} \right),\left( {6;4} \right),\left( {6;5} \right),\left( {6;6} \right)\} }
\end{array}\)

Không gian mẫu có 36 phần tử.

b) Ta có:

\(\begin{array}{l}
{\Omega _A} = \{ \left( {1;1} \right),\left( {1;2} \right),\left( {1;3} \right),\left( {1;4} \right),\\
\begin{array}{*{20}{l}}
{\left( {1;5} \right),\left( {1;6} \right),\left( {2;1} \right),\left( {2;2} \right),\left( {2;3} \right),}\\
{\left( {2;4} \right),\left( {2;5} \right),\left( {3;1} \right),\left( {3;2} \right),\left( {3;3} \right),}\\
{\left( {3;4} \right),\left( {4;1} \right),\left( {4;2} \right),\left( {4;3} \right),\left( {5;1} \right),}\\
{\left( {5;2} \right),\left( {6;1} \right)\} }
\end{array}
\end{array}\)

Tập Ω_A có 21 phần tử.

Vậy \(P\left( A \right) = \frac{{21}}{{36}} = \frac{7}{{12}}\)

c)

\(\begin{array}{l}
{\Omega _B} = \{ \left( {6;1} \right),\left( {6;2} \right),\left( {6;3} \right),\\
\left( {6;4} \right),\left( {6;5} \right),\left( {6;6} \right),\left( {1;6} \right),\\
\begin{array}{*{20}{l}}
{\left( {2;6} \right),\left( {3;6} \right),\left( {4;6} \right),\left( {5;6} \right)\} }
\end{array}
\end{array}\)

Tập Ω_B có 11 phần tử.

Vậy \(P\left( B \right) = \frac{{11}}{{36}}\)

\(\begin{array}{l}
{\Omega _C} = \{ \left( {6;1} \right),\left( {6;2} \right),\left( {6;3} \right),\left( {6;4} \right),\\
\begin{array}{*{20}{l}}
{\left( {6;5} \right),\left( {1;6} \right),\left( {2;6} \right),\left( {3;6} \right),\left( {4;6} \right),}\\
{\left( {5;6} \right)\} }
\end{array}
\end{array}\)

Vậy Ω_C có 10 phần tử.

Do đó \(P\left( C \right) = \frac{{10}}{{36}} = \frac{5}{{18}}\).

-- Mod Toán 11