Suppose there is a box with $N$ balls, $k$ white and $N-k$ black.
- If after choosing a ball it gets returned to the box, then: $p(\text{white})=\frac{k}{N},\space\space\space p(\text{black}) = \frac{N-k}{N} = 1-p(\text{white})$ — probabilities of choosing one white (black) ball, at each moment, regardless of how many balls will be selected.
What if after choosing a ball, it will be kept outside the box, and the process of selecting balls will be continued, untill all balls are selected? I am aware that this is basic, course question. I am hoping for an answer or reference, to get a level of certainty.
A third method of selecting balls is possible: suppose there is a simple device – a metal "matrix" with $N$ holes, in the shape of the box. The matrix is put on top of the box, which is then turned upside down. Also, we somehow ensure, that each hole will pass through exactly one ball. Then:
- Process is similar to case 2., because balls are not returned to box after choosing.
- Probability that given cell in the matrix will pass through white (black) ball is the same as in case 1., the basic case.
- This means, that it is not the act of not returning ball that influences probabilities in case 2. – it is the act of receiving information about ball chosen, and thus missing (in case 2.) in the box.
- Without recording the information, probabilities of case 2. are the same as in case 1.?