MHB What Is the Probability of Three Girls Given the Youngest Is Female?

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The probability of having three girls in a family of four children, given that the youngest is female, is calculated to be 0.375. This is derived from considering the possible combinations of the other three children, where the youngest is already established as a girl. The calculations show that there are three favorable outcomes (GGGB, GGBG, GBGG) out of eight total possibilities. Additionally, the discussion touches on how to determine positive correlation between two variables, although this topic is less explored. The consensus on the probability calculation provides clarity on the problem.
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i got the answer to the following problem wrong:
"there are four children in in the family. what is the probability that there are three girls, given that the youngest child is female?"

my (updated) answer:
the youngest is female, so three out of two children must be female. there are three ways of this happening, (=3C2) so, the answer is (1/2) ^3 * 3 = .375

also, how do i determine if two variables are positively correlated?
 
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Re: two probabilities question

Yuuki said:
i got the answer to the following problem wrong:
"there are four children in in the family. what is the probability that there are three girls, given that the youngest child is female?"

my (updated) answer:
the youngest is female, so three out of two children must be female. there are three ways of this happening, (=3C2) so, the answer is (1/2) ^3 * 3 = .375

also, how do i determine if two variables are positively correlated?

Lets suppose that the probability of child male and child female is the same, i.e. $p = \frac{1}{2}$. If no information is allowable, then the probability to have three girls and one boy is... $\displaystyle P = \binom{4}{3}\ \frac{1}{16} = \frac{1}{4}\ (1)$

However if You know a priori that one is famale, the probability to have three girls and one boy is the probability to have two girls and one boy among the remaining childs and it is... $\displaystyle P = \binom {3}{2}\ \frac{1}{8} = \frac{3}{8}\ (2)$ Kind regards $\chi$ $\sigma$
 
Re: two probabilities question

Here is another way to do this: writing "G" for "girl", "B" for "boy", in order from youngest to oldest we could have
GGGG
GGGB
GGBG
GGBB
GBGG
GBGB
GBBG
GBBB
The first letter is always "G" because we are told that the youngest child is a girl. The others have 2^3= 8 possible orders giving 8 possible situations. Of those 8, exactly three have 3 "G" (GGGB,, GGBG, GBGG). Assuming that boys and girls are equally likely the probability of "three girls" is 3/8= 0.375.

(If the problem were "at least three girls" we would include "GGGG" so the probability would be 4/8= 0.5.)
 
Re: two probabilities question

thanks, I'm cleared now :)
 
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