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:

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{1 9 \cdot ( 1 9 - 1 ) \cdot ( 1 9 - 2 )}{3 1 \cdot ( 3 1 - 1 ) \cdot ( 3 1 - 2 )} = 21.5573 %

p_{2} = 100 \cdot \frac{d \cdot ( d - 1 ) \cdot ( d - 2 )}{s \cdot ( s - 1 ) \cdot ( s - 2 )} = 100 \cdot \frac{1 2 \cdot ( 1 2 - 1 ) \cdot ( 1 2 - 2 )}{3 1 \cdot ( 3 1 - 1 ) \cdot ( 3 1 - 2 )} = \frac{4 4 0 0}{8 9 9} % = 4.8943 %

C_{2} (19) = (2 1 9) = \frac{1 9 !}{2 ! ( 1 9 - 2 ) !} = \frac{1 9 \cdot 1 8}{2 \cdot 1} = 171 C_{1} (12) = (1 1 2) = \frac{1 2 !}{1 ! ( 1 2 - 1 ) !} = \frac{1 2}{1} = 12 C_{3} (31) = (3 3 1) = \frac{3 1 !}{3 ! ( 3 1 - 3 ) !} = \frac{3 1 \cdot 3 0 \cdot 2 9}{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 1 9 \cdot ( 1 9 - 1 ) \cdot 1 2}{3 1 \cdot ( 3 1 - 1 ) \cdot ( 3 1 - 2 )} ≐ 45.6507 % Verifying Solution: p_{33} = 100 \cdot \frac{( 2 c ) \cdot ( 1 d )}{( 3 s )} = 100 \cdot \frac{1 7 1 \cdot 1 2}{4 4 9 5} ≐ 45.6507 %

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