Boys and girls

There are eight boys and nine girls in the class. There were six children on the trip from this class. What is the probability that left

a) only boys
b) just two boys

Result

p1 =  0.226 %
p2 =  28.507 %

Solution:

C6(17)=(176)=17!6!(17−6)!=17⋅16⋅15⋅14⋅13⋅126⋅5⋅4⋅3⋅2⋅1=12376 C6(8)=(86)=8!6!(8−6)!=8⋅72⋅1=28  p1=100⋅ (86)(8+96)=100⋅ 28(8+96)≐50221≐0.2262≐0.226%C_{{ 6}}(17) = \dbinom{ 17}{ 6} = \dfrac{ 17! }{ 6!(17-6)!} = \dfrac{ 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 } { 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 12376 \ \\ C_{{ 6}}(8)=\dbinom{ 8}{ 6}=\dfrac{ 8! }{ 6!(8-6)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ { { 8 } \choose 6 } }{ { { 8+9 } \choose 6 } }=100 \cdot \ \dfrac{ 28 }{ { { 8+9 } \choose 6 } } \doteq \dfrac{ 50 }{ 221 } \doteq 0.2262 \doteq 0.226 \%
C4(9)=(94)=9!4!(9−4)!=9⋅8⋅7⋅64⋅3⋅2⋅1=126 C2(8)=(82)=8!2!(8−2)!=8⋅72⋅1=28  p2=100⋅ (82)⋅ (96−2)(8+96)=100⋅ 28⋅ (96−2)(8+96)≐6300221≐28.5068≐28.507%C_{{ 4}}(9) = \dbinom{ 9}{ 4} = \dfrac{ 9! }{ 4!(9-4)!} = \dfrac{ 9 \cdot 8 \cdot 7 \cdot 6 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 126 \ \\ C_{{ 2}}(8)=\dbinom{ 8}{ 2}=\dfrac{ 8! }{ 2!(8-2)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ p_{2}=100 \cdot \ \dfrac{ { { 8 } \choose 2 } \cdot \ { { 9 } \choose 6-2 } }{ { { 8+9 } \choose 6 } }=100 \cdot \ \dfrac{ 28 \cdot \ { { 9 } \choose 6-2 } }{ { { 8+9 } \choose 6 } } \doteq \dfrac{ 6300 }{ 221 } \doteq 28.5068 \doteq 28.507 \%

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