Imagine two people got married to each other and had two children. If one is a girl, what is the probability that the other child is also a girl? Instinctively, you should respond “1/2!” and you’re right. Except… you’re not.
Think of all the possible ways that a girl could be in the family. There’s the two girls (gg), then the two possible mix of both (gb and bg). That means that math is biased and there’s only a 1/3 chance that the other child will be a girl as well.
But hold on. The 1/3 only applies if gender in families is dependent, or one child’s gender is dependent on another.
You see, you are made of 23 pairs of chromosomes, each pair fuelled by your parents’ chromosomes. For the gender chromosome, it consists of X and Y. Here Y is the dominant chromosome, meaning that if a person is to have a Y chromosome, they become a male. Now, assuming there is no same-sex marriage, children have the gender chromosome of either XY, YX or XX (XX is the female).
Now, for the final answer: gender is an independent variable, meaning whatever the gender of the first child is, the chance of the second child being a girl is 1/2.
Why? If we take a look at the scenarios, it would be:
|Possibilities||Child one||Child two|
Now, in the puzzle, it has been stated that one of the children is a girl. If we let the stated child be child one, then it comes down to two scenarios: GB and GG. Two scenarios, and if the probability of being a boy or girl is 50/50, then the chance of the second child being a girl is 50%, or one in two.
More simply, since the genders of one child based on another is independent, the chance is not affected by the first child, therefore it is still 50%, no matter what. This could be done with four children, seven children, 20 children like Maria and James Burton, or more than 860 like Ismail Ibn Sharif, who registered 525 boys and 342 girls with many, many women. Always 50%.
From what I know.
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Reblogged this on Love & Love Alone.