Question Number 165673 by mkam last updated on 06/Feb/22
Answered by mr W last updated on 06/Feb/22
![to select 3 from 10 students there are totally C_3 ^(10) =120 possibilties. 0 girl: C_3 ^6 =20 possibilities, p(0)=((20)/(120)) 1 girl: C_1 ^4 C_2 ^6 =60, p(1)=((60)/(120)) 2 girls: C_2 ^4 C_1 ^6 =36, p(2)=((36)/(120)) 3 girls: C_3 ^4 =4, p(3)=(4/(120)) E[X]=0×((20)/(120))+1×((60)/(120))+2×((36)/(120))+3×(4/(120)) =1.2 Var(X)=(0−1.2)^2 ×((20)/(120))+(1−1.2)^2 ×((60)/(120))+(2−1.2)^2 ×((36)/(120))+(3−1.2)^2 ×(4/(120)) =0.56 or E[X^2 ]=0^2 ×((20)/(120))+1^2 ×((60)/(120))+2^2 ×((36)/(120))+3^2 ×(4/(120)) =2 Var(X)=E(X^2 )−(E(X))^2 =2−1.2^2 =0.56](https://www.tinkutara.com/question/Q165700.png)
$${to}\:{select}\:\mathrm{3}\:{from}\:\mathrm{10}\:{students}\:{there} \\ $$$${are}\:{totally}\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120}\:{possibilties}. \\ $$$$\mathrm{0}\:{girl}:\:{C}_{\mathrm{3}} ^{\mathrm{6}} =\mathrm{20}\:{possibilities},\:{p}\left(\mathrm{0}\right)=\frac{\mathrm{20}}{\mathrm{120}} \\ $$$$\mathrm{1}\:{girl}:\:{C}_{\mathrm{1}} ^{\mathrm{4}} {C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{60},\:{p}\left(\mathrm{1}\right)=\frac{\mathrm{60}}{\mathrm{120}} \\ $$$$\mathrm{2}\:{girls}:\:{C}_{\mathrm{2}} ^{\mathrm{4}} {C}_{\mathrm{1}} ^{\mathrm{6}} =\mathrm{36},\:{p}\left(\mathrm{2}\right)=\frac{\mathrm{36}}{\mathrm{120}} \\ $$$$\mathrm{3}\:{girls}:\:{C}_{\mathrm{3}} ^{\mathrm{4}} =\mathrm{4},\:{p}\left(\mathrm{3}\right)=\frac{\mathrm{4}}{\mathrm{120}} \\ $$$${E}\left[{X}\right]=\mathrm{0}×\frac{\mathrm{20}}{\mathrm{120}}+\mathrm{1}×\frac{\mathrm{60}}{\mathrm{120}}+\mathrm{2}×\frac{\mathrm{36}}{\mathrm{120}}+\mathrm{3}×\frac{\mathrm{4}}{\mathrm{120}} \\ $$$$=\mathrm{1}.\mathrm{2} \\ $$$${Var}\left({X}\right)=\left(\mathrm{0}−\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} ×\frac{\mathrm{20}}{\mathrm{120}}+\left(\mathrm{1}−\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} ×\frac{\mathrm{60}}{\mathrm{120}}+\left(\mathrm{2}−\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} ×\frac{\mathrm{36}}{\mathrm{120}}+\left(\mathrm{3}−\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} ×\frac{\mathrm{4}}{\mathrm{120}} \\ $$$$=\mathrm{0}.\mathrm{56} \\ $$$$ \\ $$$${or} \\ $$$${E}\left[{X}^{\mathrm{2}} \right]=\mathrm{0}^{\mathrm{2}} ×\frac{\mathrm{20}}{\mathrm{120}}+\mathrm{1}^{\mathrm{2}} ×\frac{\mathrm{60}}{\mathrm{120}}+\mathrm{2}^{\mathrm{2}} ×\frac{\mathrm{36}}{\mathrm{120}}+\mathrm{3}^{\mathrm{2}} ×\frac{\mathrm{4}}{\mathrm{120}} \\ $$$$=\mathrm{2} \\ $$$${Var}\left({X}\right)={E}\left({X}^{\mathrm{2}} \right)−\left({E}\left({X}\right)\right)^{\mathrm{2}} =\mathrm{2}−\mathrm{1}.\mathrm{2}^{\mathrm{2}} =\mathrm{0}.\mathrm{56} \\ $$
Commented by Tawa11 last updated on 06/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by peter frank last updated on 09/Feb/22
$$\mathrm{nice} \\ $$