Thể tích khối tròn xoay tạo thành khi quay hình phẳng \(\left( H \right)\) giới hạn bởi các đường \(y = f\left( x \right);y = 0;x = a;x = b\,\,\left( {a < b} \right)\) quanh trục hoành bằng:    

Xét \(\Delta SBA\) có 3 điểm \({\rm{M}},{\rm{N}},{\rm{H}}\) thẳng hàng. Theo định lý Menelaus ta có:

\(\frac{{MS}}{{MA}} \cdot \frac{{HA}}{{HB}} \cdot \frac{{NB}}{{NS}} = 1 \Rightarrow \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{{HA}}{{HB}} = 1 \Rightarrow \frac{{HA}}{{HB}} = 4 \Rightarrow HA = 4HB\).

Ta có: \(\overrightarrow {MH}  = \overrightarrow {MA}  + \overrightarrow {AB}  + \overrightarrow {BH}  = \frac{2}{3}\overrightarrow {SA}  + \overrightarrow {AB}  + \frac{1}{5}\overrightarrow {AB}  = \frac{2}{3}\overrightarrow {SA}  + \frac{6}{5}\overrightarrow {AB} \). Chọn B.

Ta có:

\({\rm{\Omega }} = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {2,1} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,4} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right)} \right\}\).

\(A = \left\{ {\left( {2,1} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right)} \right\}\).

\(B = \left\{ {\left( {2,1} \right),\left( {3,1} \right),\left( {4,1} \right),\left( {1,3} \right),\left( {2,3} \right),\left( {4,3} \right)} \right\}\).

\(A \cap B = \left\{ {\left( {2,1} \right),\left( {2,3} \right),\left( {4,1} \right),\left( {4,3} \right)} \right\}\).

Ta có: \(n\left( {\rm{\Omega }} \right) = 12\).

\(n\left( A \right) = 2.3 = 6 \Rightarrow P\left( A \right) = \frac{6}{{12}} = \frac{1}{2}\).

\(n\left( {A \cap B} \right) = 4 \Rightarrow P\left( {A \cap B} \right) = \frac{4}{{12}} = \frac{1}{3}\)

Vậy \(P\left( {B\mid A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \frac{{\frac{1}{3}}}{{\frac{1}{2}}} = \frac{2}{3}\). Chọn C.