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 one girl?

Correct answer:

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

Step-by-step explanation:

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

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

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

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