Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.
What is the probability that only girls will be elected?
What is the probability that only boys will be elected?
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(1) What is the probability that only girls will be elected?Just for practice, since the most efficient way to solve this problem is the Probability approach.
(1) probability approach:\(\frac{4}{10} * \frac{3}{9} * \frac{2}{8} =
\frac{1}{30}\)But I need to practice so I solved this problem with the other 3 approaches:
(2) Reversal Probability approach: P = 1 - q (q=probability of not picking only girls, this means, picking 1, 2 or 3 boys):
--> 1 boy + 2 girls = BGG (and combinations):
\(1 - \frac{6}{10} * \frac{4}{9} * \frac{3}{8} * \frac{3!}{2!} = \frac{3}{10}\)
-->2 boys + 1 Girl = BBG (And combinations):
\(\frac{6}{10} *\frac{5}{9} * \frac{4}{8} * \frac{3!}{2!} = \frac{1}{2}\)
--> 3 boys:
\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3!}{3!} = \frac{1}{6}\)
Then, we add all the probabilities:
\(\frac{3}{10} + \frac{1}{2}+ \frac{1}{6} = \frac{29}{30} = q\)
\(1-q = 1- \frac{29}{30} =
\frac{1}{30}\) = P(3) Combinatorial approach:\(\frac{C^4_3}{
C^10_3} = \frac{4}{120} =
\frac{1}{30}\)(4) Reversal combinatorial approach: \(1 - \frac{{C^6_1 * C^4_2 + C^6_2 * C^4_1 + C^6_3}}{C^10_3} = 1- \frac{{36 + 60 + 20}}{120} = 120/120 - \frac{116}{120} =
\frac{1}{30}\)(2) What is the probability that only boys will be elected?(1) Probability approach:\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} = \frac{4}{24} = \frac{1}{6}\)
(2) Reversal probability approach --> P = 1 - q
1girl + 2 boys: \(\frac{4}{10} *\frac{6}{9} *\frac{5}{8} * \frac{3!}{2!} = \frac{1}{6} * 3 = \frac{1}{2}\)
2 girls + 1 boy: \(\frac{4}{10} *\frac{3}{9} * \frac{6}{8} *\frac{3!}{2!} = \frac{1}{10} * 3 = \frac{3}{10}\)
3 girls: \(\frac{4}{10} *\frac{3}{9} * \frac{2}{8} = \frac{1}{30}\)
(3) Combinatorial approach:\(\frac{C^6_3}{C^10_3} = \frac{20}{120} = \frac{1}{6}\)
(4) Reversal Comb approach:\(\frac{C^4_1 * C^6_2 + C^4_2 * C^6_1 + C^4_3}{C^10_3}\)
1 girl + 2 boys: \(C^4_1 * C^6_2 = 60\)
2 girls + 1 boy: \(C^4_2 * C^6_1 = 36\)
3 girls:\(C^4_3 = 4\)
60 + 36 + 4 = 100
\(\frac{120}{120} - \frac{100}{120} = \frac{20}{120} = \frac{1}{6}\)
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68% (02:30) correct
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