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1. Chim không ăn kẹo
2. Xin địa chỉ có bò để thống kê tính toán
a. Xác suất có 1 con trai:
\(C_3^1.0,53^1.\left(1-0,53\right)^2=0,351231\)
b. Xác suất sinh không quá 1 con trai:
\(1-C_3^3.0,53^3.\left(1-0,53\right)^0=0,851123\)
c. Số trai nhiều hơn gái (3 trai 0 gái, 2 trai 1 gái):
\(C_3^3.0,53^3.\left(1-0,53\right)^0+C_3^2.0,53^2.\left(1-0,53\right)^1=0,544946\)
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1b/ \(\overrightarrow{AB}=\left(1;1;3\right);\overrightarrow{u_{Oy}}=\left(0;1;0\right)\)
Vì \(\left(P_2\right)//AB//Oy\Rightarrow\overrightarrow{n_{\left(P_2\right)}}=\left[\overrightarrow{AB},\overrightarrow{u_{Oy}}\right]=\left(\left|\begin{matrix}1&3\\1&0\end{matrix}\right|,\left|\begin{matrix}3&1\\0&0\end{matrix}\right|,\left|\begin{matrix}1&1\\0&1\end{matrix}\right|\right)=\left(-3;0;1\right)\)
\(\Rightarrow\left(P_2\right):-3\left(x+3\right)+z-5=0\Leftrightarrow\left(P_2\right):3x-z+14=0\)
2b/
\(\overrightarrow{u_{Ox}}=\left(1;0;0\right);\overrightarrow{n_{\left(Q\right)}}=\left(3;2;5\right)\)
\(\Rightarrow\overrightarrow{n_{\left(\beta\right)}}=\left[\overrightarrow{u_{Ox}},\overrightarrow{n_{\left(Q\right)}}\right]=\left(0;-5;2\right)\)
\(d\left(O,\left(\beta\right)\right)=\dfrac{\left|d\right|}{\sqrt{25+4}}=\sqrt{29}\Rightarrow d=\pm29\)
\(\Rightarrow\left[{}\begin{matrix}\left(\beta\right):-5y+2z+29=0\\\left(\beta\right):-5y+2z-29=0\end{matrix}\right.\)
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