Classroom

Of the 26 pupils in the classroom, 12 boys and 14 girls, four representatives are picked to the odds of being:
a) all the girls
b) three girls and one boy
c) there will be at least two boys

Result

p1 =  0.067
p2 =  0.292
p3 =  0.641

Solution:

p1=14 13 12 1126 25 24 23=7711500.067=0.067p_{ 1 } = \dfrac{ 14 \cdot \ 13 \cdot \ 12 \cdot \ 11 }{ 26 \cdot \ 25 \cdot \ 24 \cdot \ 23 } = \dfrac{ 77 }{ 1150 } \doteq 0.067 = 0.067
C3(14)=(143)=14!3!(143)!=141312321=364  C1(12)=(121)=12!1!(121)!=121=12  C4(26)=(264)=26!4!(264)!=262524234321=14950  p2=(143) (121)/(264)=364 12/14950=1685750.2922=0.292C_{{ 3}}(14) = \dbinom{ 14}{ 3} = \dfrac{ 14! }{ 3!(14-3)!} = \dfrac{ 14 \cdot 13 \cdot 12 } { 3 \cdot 2 \cdot 1 } = 364 \ \\ \ \\ C_{{ 1}}(12) = \dbinom{ 12}{ 1} = \dfrac{ 12! }{ 1!(12-1)!} = \dfrac{ 12 } { 1 } = 12 \ \\ \ \\ C_{{ 4}}(26) = \dbinom{ 26}{ 4} = \dfrac{ 26! }{ 4!(26-4)!} = \dfrac{ 26 \cdot 25 \cdot 24 \cdot 23 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 14950 \ \\ \ \\ p_{ 2 } = { { 14 } \choose 3 } \cdot \ { { 12 } \choose 1 } / { { 26 } \choose 4 } = 364 \cdot \ 12 / 14950 = \dfrac{ 168 }{ 575 } \doteq 0.2922 = 0.292
p3=1p1p2=10.0670.2922=6411000=0.641p_{ 3 } = 1-p_{ 1 }-p_{ 2 } = 1-0.067-0.2922 = \dfrac{ 641 }{ 1000 } = 0.641



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