Note: Cheng is under the impression that the answer has to be 50%, therefore making Cheng a 33% denier.
Cheng is bored, so chengameepheus has decided to use ninnish examples to exemplify what to do. In my first attempt, i’ll use a Presicion Chengsicion Tree, not to be confused with an exacto tree (snipping would be innappropriate for this example).
A bungalow(100) of children evenly distributed between male and female
One family….with 2 children….
if first pick is male…49 males are left and 50 females are left
if first pick is female…49 females and 50 males are left
We take the if male thing, thats 49/99 that the second is a male. Since we want infinite (insert any number here), we take infinite divided by 2(since it is 50% prob its male) minus 1(for the male we took out), then divide it by infinite divided by 2(since thats the amount of possible females left) + the numerator.
(infinite/2-1) + (infinite/2)
Thats so close to 50% that Cheng won’t do the smath. Cheng’s precision chengcision tree has declared 50%.
Cheng now will call into question the prob. tree that everyone is making. This thing
P(X = 0) = 0.25
P(X = 1) = 0.50
P(X = 2) = 0.25
(taken from gas’s thing)
The situation we are using is not as simple as it looks. I will also assign F (female) or M (Male). In a normal plot, its MM, FF, FM, MF. This overlooks the fact that we aren’t just combining FM and FM. This means I will assign a 1 and 2 to the letter combos also. This is because we don’t know whether the male is the first or second child. The number is to declare whether its the first or second of its kind to be chosen. Here are all the possibilities.
Once we chrisliminate the (f1,f2) as well as the (f2, f1), the results become apparent. It becomes (M1, M2) + (M2, M1) + (M1,F1) + (F1, M1) divided by (M1,F1) + (F1, M1), also known as 4 divi, oh shintos, i’m dyslexic or something. switch those around, then its also known as 2 divided by 4 = .5 or 50%. Please attempt a refute job on this, its just a ponderoso coming from chengstein’s thinking muscle.