Say for instance you have 4 possible items of type $a$,$b$,$c$ and $d$ in a bag. You know there are $100$ items in the bag. You don't know how many of each are in the bag. The items are not put back in the bag.
- Am I correct in saying before you draw any out of the bag, the probability is $\frac{25}{100}$ for each?
- If you then draw an items of type '$a$' out of then bag, how does this effect the probabilities?
- What if you then draw out an item of type '$b$'?
Thanks
Edit: I think i'm looking for something like Josu Etxezarreta Martinez answer.
Additional follow up questions:
does $P(a)= \frac{\#a-1}{99}$ mean you are more likely to draw an 'a' on the next draw compared to a 'd' when $P(d)= \frac{\#d-1}{99}$?
if on a third iteration I again selected an 'a', would that make $P(a)= \frac{\#a-2}{97}$?
For the first question you are assuming that there are 25 elements of each of the items in the bag. The actualk real distribution would be:
$P(x_i)=\frac{\#x_i}{100}$ where $\sum_i\#x_i=100$, and where $\#$ denotes number of elements.
For the second and third question you must use the same method, but when taking out an item of an specific kind you should rest one to the number of items in that specific group and also one to the total amount of elements in the bag, that would be
(b) $P(a)= \frac{\#a-1}{99}$ $P(b)= \frac{\#b}{99}$ $P(c)= \frac{\#c}{99}$ $P(d)= \frac{\#d}{99}$
(c) $P(a)= \frac{\#a-1}{98}$ $P(b)= \frac{\#b-1}{98}$ $P(c)= \frac{\#c}{98}$ $P(d)= \frac{\#d}{98}$