This is likely a question with answers everywhere, but I am unsure how to phrase it so I find answers.
Lets say you have $100$ hammers where $23$ are defective.
Example 1: If you select two, one at a time, without replacement, what is the probability the second one is defective?
A1: The way I would naturally answer it is:
$\frac{23}{100}\times\frac{22}{99}+\frac{77}{100}\times\frac{23}{99} = \frac{23}{100}$.
This is the same probability that the first hammer chosen being defective.
Example 2: If you select $3$, one at a time, without replacement, what is the probability the first is defective but the last is not?
A1: Once again the way I would naturally answer it is:
$\frac{23}{100}\times\frac{22}{99}\times\frac{77}{98}+\frac{23}{100}\times\frac{77}{99}\times\frac{76}{98} = \frac{161}{900} = \frac{23}{100}\times\frac{77}{99}$.
This is the same probability that the first hammer chosen is defective, and the second not.
What I am getting at is the probabilites were the same when ignoring the "does not matter" choice. Assuming I am not mistaking something, how does this make sense? I can't seem to wrap my head around it intuitively or mathmatically. I would expect that because it is without replacement, even a "does not matter" changes the probability by reducing the total hammer count.
