Sampling with and without replacement.

Question

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 $?$

Answer 1

This is obviously true for the first draw

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