# Sampling with and without replacement.

Prove that the probability of drawing a unit at any draw from a population of size N, remains same in without and with replacement sampling scheme.

I know how to prove that this probability is $$\frac{1}{N}$$ for without replacement. But how should I prove that the probability is same for with replacement $$?$$

For the $$k+1$$th draw without replacement, with $$k \lt N$$
• The probability that the first $$k$$ draws did not contain the item was $$\dfrac{n-1}{n} \times \dfrac{n-2}{n-1} \times \cdots \times \dfrac{n-k}{n-k+1} = \dfrac{n-k}{n}$$
• Conditional on that, the probability that the $$k+1$$th draw from the remaining $$n-k$$ undrawn was the item would be $$\dfrac{1}{n-k}$$
• So the marginal probability at the start that the $$k+1$$th draw from all the $$n$$ was the item would be $$\dfrac{n-k}{n}\times \dfrac{1}{n-k} = \dfrac{1}{n}$$ as expected
There is also a symmetry argument: if you make $$n$$ draws without replacement, then the problem is equivalent to positioning the desired item in $$n$$, and these positions are equally likely