calculating probability of a kid have a sister Suppose there is equal probability of Boy (B) and Girl (G). If a family has 3 kids, wondering what is the probability of a kid has a sister in the same family.
I have two methods of calculation, first method is a bit naive, and 2nd method is faster calculation, and the question is for the 2nd method, I am not sure if under all scenarios the probability of two remaining kids are both boys could be calculated as 1/2 * 1/2 = 1/4. Thanks.
Method 1,


*

*G G G (1/8 probability have 3 boys in a family), in this condition,
probability will be 1 for any kid to have a sister

*B B B (1/8 probability have 3 girls in a family) in this condition, probability
will be 0 for any kid to have a sister

*G B B (3/8 probability have 2 boy2 and 1 girls in a family) in this condition, probability will be 2/3 for any kid to have a sister

*G G B (3/8 probability have 2 girls and 1 boy in a family) in this condition, probability will be 1 for any kid to have a sister


so probability of a kid having a sister is
1/8 + 0 + 1/4 + 3/8 = (1+2+3)/8 = 3/4
Method 2,
Suppose we select any kid in a family, and for the other two kids in the same family, the probability of two boys are 1/4 (when there is no sister scenario, my question is whether the probability is correct calculate in this way, according to 4 scenarios of family), so the probability of a kid having a sister is 1 - 1/4 = 3/4, the same result of method 1.
thanks in advance,
Lin
 A: Both methods are fine. You didn't really specify the problem, but I gather you meant to imply that a child is selected from the family uniformly randomly, and that the genders of the three children are independent. In that case, both your arguments are correct.
A: Method 1: Using the Law of Total Probability.
Let $S$ be the event of selecting a child with a sister.  Let $F_{\rm GGG}, F_{\rm GGB}, F_{\rm GBB}, F_{\rm BBB}$ be the event of selecting a child from the relevant family structure; given that we know we are selecting from families with three children.
The count of girls in a family has a Binomial Distribution.   And for any given number of girls in a three child family, the probability of selecting a child with a sister is easily determined.   So, as you calculated for method 1. $$\begin{align}\mathsf P(S) =&~ \mathsf P(S\mid F_{\rm GGG})\mathsf P(F_{\rm GGG})+\mathsf P(S\mid F_{\rm GGB})\mathsf P(F_{\rm GGB})+\mathsf P(S\mid F_{\rm GBB})\mathsf P(F_{\rm GBB})+\mathsf P(S\mid F_{\rm BBB})\mathsf P(F_{\rm BBB})\\[1ex]=&~ \mathsf P(S\mid F_{\rm GGG})\cdot\tfrac 1 8+\mathsf P(S\mid F_{\rm GGB})\cdot\tfrac 38+\mathsf P(S\mid F_{\rm GBB})\cdot\tfrac 38+\mathsf P(S\mid F_{\rm BBB})\cdot\tfrac 18\\[1ex]=&~ 1\cdot \tfrac 1 8 ~+~ 1\cdot\tfrac 38~+~\tfrac 2 3\cdot \tfrac 3 8~+~0\cdot \tfrac 18 \\[1ex]=&~ \tfrac 3 4\end{align}$$
Method 2:
The probability that at least one, of the two children that were not selected, was a girl is: $\mathsf P(S)=\tfrac 3 4$, as you determined using the law of complements.

To address you concerns.   $\mathsf P(S\mid F_{\rm GBB}) = \tfrac 23$ because in such a family the probability of selecting a child with a sister is the probability of selecting one of the two boys.   If you select a boy from a family with two boys and one girl, then he will have a sister, if you select the girl from that family then she will not.
And so on for the other family types, where fixing the number of girls among the three children determines how many children have a sister.   In a family with two or more girls, every child has a sister, and in a family with three boys, none will.
Where as when selecting a family (of three children) at random, there isn't a fixed number of girls, and so each child independently has a probability of $\tfrac 12$ for being a girl; and any child so selected has $\tfrac 34$ probability of having a sister.
