\(n\left(\Omega\right)=C^4_{12}\)
a) 4 Hs lớp A : \(C^4_5\)
4 Hs lớp B : 1
P = \(\frac{C^4_5+1}{C^4_{12}}=\frac{2}{165}\)
b) 4 Hs thuộc cả 3 lớp : \(C^2_5\cdot4\cdot3+5\cdot C^2_4\cdot3+5\cdot4\cdot C^2_3\)
P = 1 - \(\frac{C^2_5\cdot4\cdot3+5\cdot C^2_4\cdot3+5\cdot4\cdot C^2_3}{C^4_{12}}=\frac{5}{11}\)