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

Correct result:

p₁ =  0.067
p₂ =  0.2922
p₃ =  0.6409

Solution:

$p_{1}=\dfrac{ 14 \cdot \ 13 \cdot \ 12 \cdot \ 11 }{ 26 \cdot \ 25 \cdot \ 24 \cdot \ 23 }=\dfrac{ 77 }{ 1150 }=0.067$

$$\begin{gathered} C_3(14)=\left({\displaystyle \genfrac{}{}{0px}{}{14}{3}}\right)={\displaystyle \frac{14!}{3!(14-3)!}}={\displaystyle \frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}}=364 \\ {C}_1(12)=\left({\displaystyle \genfrac{}{}{0px}{}{12}{1}}\right)={\displaystyle \frac{12!}{1!(12-1)!}}={\displaystyle \frac{12}{1}}=12 \\ {C}_4(26)=\left({\displaystyle \genfrac{}{}{0px}{}{26}{4}}\right)={\displaystyle \frac{26!}{4!(26-4)!}}={\displaystyle \frac{26\cdot 25\cdot 24\cdot 23}{4\cdot 3\cdot 2\cdot 1}}=14950 \\ {p}_2=\left(\genfrac{}{}{0px}{}{14}{3}\right)\cdot \left(\genfrac{}{}{0px}{}{12}{1}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{26}{4}\right)=364\cdot 12\mathrm{/}14950={\displaystyle \frac{168}{575}}=0.2922 \end{gathered}$$

$p_3=1-p_1-p_2=1-0.067-0.2922={\displaystyle \frac{737}{1150}}=0.6409$

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