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I know the problem has many answer in this site or other sites. But first idea of mine is using Permutation And Combination. Here is the problem:

Have known family of three children, one of whom is a girl, asks for the probability of at least one boy.

First of all, I want to use three boxes. I'm assuming that X event means one of whom is a girl. So I pull one box out of three which will be put in a girl, the other boxes will be put in a girl or a boy. Therefore the formula is $${ A }_{ 3 }^{ 1 }+{ A }_{ 3 }^{ 2 }{ C }_{ 2 }^{ 1 }=3+3\cdot 2\cdot 2=15$$ I’m assuming that Y event means one of whom is a girl and at least one boy. So I pull one box out of three which will be put in a girl and one box which will be put in a boy, the other one will be put in a girl or a boy. Therefore the formula is $${ A }_{ 3 }^{ 1 }+{ A }_{ 3 }^{ 1 }{ +C }_{ 2 }^{ 1 }=3+3+2=8$$ Therefore, I think result is $$\frac { { A }_{ 3 }^{ 1 }+{ A }_{ 3 }^{ 1 }{ +C }_{ 2 }^{ 1 } }{ { A }_{ 3 }^{ 1 }+{ A }_{ 3 }^{ 2 }{ C }_{ 2 }^{ 1 } }=\frac { 8 }{ 15 } $$ but it isn’t correct obviously. What is wrong with my solution?

Boris
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    Question is very unclear. Please try to make a clearer explanation of your reasoning and the problem. – Sofía Marlasca Aparicio Sep 21 '17 at 13:57
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    @marlasca23 I believe the problem is this: Consider a family with three children, one of whom is a girl. What is the probability that there is at least one boy? But yes, the rest is indeed unclear. What does the notation $A_3^1$ mean? – Théophile Sep 21 '17 at 14:11
  • @marlasca23 here is a similar question In a family with three children, what are the chances, if one of the children is a girl, that at least one children is boy? – Boris Sep 21 '17 at 14:13
  • @Théophile ${ A }{ 3 }^{ 1 }=3,\quad { A }{ 3 }^{ 2 }=3\cdot 2,\quad { A }_{ 3 }^{ 3 }=3\cdot 2\cdot 1$ – Boris Sep 21 '17 at 14:16
  • @Théophilet This is a very common symbol in our Chinese textbooks – Boris Sep 21 '17 at 14:17
  • Your probability exceeds $1$. If you are counting favorable cases, you need to divide by all possible outcomes. However, the number of favorable cases you listed is greater than the number of possible outcomes in which a family can have three children, including a girl. – N. F. Taussig Sep 21 '17 at 14:18
  • @N.F.Taussig Why the number of favorable cases you listed is greater than the number of possible outcomes? – Boris Sep 21 '17 at 14:27
  • There are seven possible sequences of boys and girls in a family of three children that has at least one girl: bbg, bgb, gbb, bgg, gbg, ggb, ggg. Your numerator simplifies to $3 + 3 + 2 = 8$. – N. F. Taussig Sep 21 '17 at 14:39
  • @N.F.Taussign If the number is very large, how do you enumerate all the cases? For example, there are 1000 children in a family, 100 of them are girls, asks for the probability of at least one boy. – Boris Sep 21 '17 at 14:49

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If I understand your line of reasoning, the answer would be

$$ \frac{A_{3}^{2} + A_{3}^{1}}{A_{3}^{1}+A_{3}^{2}+ A_{3}^{3}} = \frac{3+3}{3+3+1}= \frac{6}{7} $$

where $A_{i}^{j}$ is the number of (unordered) ways of picking $j$ objects among $i$.Possible cases are either 1,2, or 3 girls, and favorables cases are 1 or 2 boys (you can't have three boys if you have a girl).

citronrose
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