Having a girl the same as AT LEAST one girl? CONFUSION!!

#1
It’s a combinatorics project for my beginner stats class. I’ve done the project several times and keep getting it wrong. The question reads: “ Is having a girl the same event as having AT LEAST one girl?” AND then it asks me: “Is it possible to calculate these probabilities?” This project is keeping me from getting a B. Please help!! :( 67E7A37F-4410-482F-B717-BA873FA7913E.png
 
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Karabiner

TS Contributor
#2
Is having 1 Dollar the same as having at least 1 Dollar?

Little Nemo (4 years old) today received his first and only Dollar, as a birthday present.

Bill Gates also has got 1 Dollar. At least.
 
#3
Is having 1 Dollar the same as having at least 1 Dollar?

Little Nemo (4 years old) today received his first and only Dollar, as a birthday present.

Bill Gates also has got 1 Dollar. At least.
So the family with 8 boys aspiring to have one girl has nothing to do with this question, correct? If so, then ( please correct me) having one girl is NOT the same event of having at least one girl.
 
#4
P( having one girl)= 1/2 (.50) or 50%
P( having at least one girl)= 1- P( no girl). We can't calculate this because there's a 50% chance it could be a boy or a girl. Does this make sense?
 

Karabiner

TS Contributor
#5
The first statement goes beyond simple probabilty calculations. Maybe it is empirically true that p(girl) is small after 8 boys in a row, because of some biological/mediacal characteristic of one or both parents?

Otherwise, I agree that one would expect just p=0.50

The second statement is true (in a sense of almost a guarantee) before the birth of the first child. p(no girl) is 0.5^9 in case of 9 children.
But the statement cannot be used for predicting the sex of the 9th child after 8 children have been born.

With kind regards

Karabiner