Created trio

What is the probability that in the created trio, which consists of 19 boys and 12 girls, they will be:
a) the boys themselves
b) the girls themselves
c) 2 boys and 1 girl?

Correct answer:

p₁ =  21.5573 %
p₂ =  4.8943 %
p₃ =  45.6507 %

Step-by-step explanation:

$$\begin{gathered} c=19 \\ d=12 \\ s=c+d=19+12=31 \\ {p}_1=100\cdot {\displaystyle \frac{c\cdot (c-1)\cdot (c-2)}{s\cdot (s-1)\cdot (s-2)}}=100\cdot {\displaystyle \frac{19\cdot (19-1)\cdot (19-2)}{31\cdot (31-1)\cdot (31-2)}}=21.5573\mathrm{\%} \end{gathered}$$

$p_2=100\cdot {\displaystyle \frac{d\cdot (d-1)\cdot (d-2)}{s\cdot (s-1)\cdot (s-2)}}=100\cdot {\displaystyle \frac{12\cdot (12-1)\cdot (12-2)}{31\cdot (31-1)\cdot (31-2)}}={\displaystyle \frac{4400}{899}}=4.8943\mathrm{\%}$

$$\begin{gathered} C_2(19)=\left({\displaystyle \genfrac{}{}{0px}{}{19}{2}}\right)={\displaystyle \frac{19!}{2!(19-2)!}}={\displaystyle \frac{19\cdot 18}{2\cdot 1}}=171 \\ {C}_1(12)=\left({\displaystyle \genfrac{}{}{0px}{}{12}{1}}\right)={\displaystyle \frac{12!}{1!(12-1)!}}={\displaystyle \frac{12}{1}}=12 \\ {C}_3(31)=\left({\displaystyle \genfrac{}{}{0px}{}{31}{3}}\right)={\displaystyle \frac{31!}{3!(31-3)!}}={\displaystyle \frac{31\cdot 30\cdot 29}{3\cdot 2\cdot 1}}=4495 \\ {p}_3=100\cdot {\displaystyle \frac{3\cdot c\cdot (c-1)\cdot d}{s\cdot (s-1)\cdot (s-2)}}=100\cdot {\displaystyle \frac{3\cdot 19\cdot (19-1)\cdot 12}{31\cdot (31-1)\cdot (31-2)}}\doteq 45.6507=45.6507\mathrm{\%} \\ \text{ Verifying Solution: } \\ {p}_{33}=100\cdot {\displaystyle \frac{\left(\genfrac{}{}{0px}{}{c}{2}\right)\cdot \left(\genfrac{}{}{0px}{}{d}{1}\right)}{\left(\genfrac{}{}{0px}{}{s}{3}\right)}}=100\cdot {\displaystyle \frac{171\cdot 12}{4495}}\doteq 45.6507\mathrm{\%} \end{gathered}$$

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