[tex]\it Pentru\ probabilitate\ folosim\ formula:\\ \\ p=\dfrac{nr.\ cazuri\ favorabile}{nr.\ cazuri\ posibile}\\ \\ Cazurile\ favorabile\ sunt:\\ \\ \{1\},\ \{3\},\ \{5\},\ \{1,\ 3\},\ \{1,\ 5\},\ \{3,\ 5\},\ \{1,\ 3,\ 5\}.\\ \\ Deci,\ avem\ 7\ cazuri\ favorabile.\\ \\ Nr.\ cazurilor\ posibile=2^6-1=63\\ \\ p=\dfrac{\ \ 7^{(7}}{63}=\dfrac{1}{9}[/tex]
5)
[tex]\it Vom\ determina\ lungimile\ laturilor \ triunghiului:\\ \\ AB=\sqrt2\\ \\ AC=\sqrt5\\ \\ BC=\sqrt13\\ \\ AB^2+AC^2=2+5=7<13=BC^2 \Rightarrow\ m(\hat A)>90^0 \Rightarrow\\ \\ \Rightarrow m(\hat C)<90^0 \Rightarrow sinC=\sqrt{1-cos^2C}\\ \\ cosC=\dfrac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\dfrac{5+13-2}{2\cdot\sqrt5\cdot\sqrt{13}}=\dfrac{16^{(2}}{2\sqrt{65}}=\dfrac{8}{\sqrt65}\Rightarrow\\ \\ \\ \Rightarrow cos^2C= \dfrac{64}{65}[/tex]
[tex]\it\ sinC=\sqrt{1-\dfrac{64}{65}}=\sqrt{\dfrac{1}{65}}=\dfrac{1}{\sqrt{65}}=\dfrac{\sqrt{65}}{65}[/tex]
