Probability of sampling with and without replacement

In sampling without replacement the probability of any fixed element in the population to be included in a random sample of size $r$ is $\frac{r}{n}$. In sampling with replacement the corresponding probability is $\left[1- \left(\frac{1}{1-n}\right)^r\right]$.

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– 6005
Sep 10 '13 at 19:06

Sampling without replacement

Just a note in terms of nomenclature: $${n \choose r} = {_n}C_r = \frac{n!}{r!(n-r)!}$$ There are ${n \choose r}$ ways to select the sample of $r$ elements from the pool of $n$ items. That is our denominator—the universe of possible results. To count the number of samples which include our special fixed element (call it $x^*$) we can just realize that we must pick $x^*$, and there is only one way to do that (${1 \choose 1}$, and that leaves us a pool of $n-1$ item from which we need to select $r-1$ to fill out the sample. There are: $${1 \choose 1}{n-1 \choose r-1}$$ ways to create our desired samples. So the probability of having $x^*$ in our mix is: $$\frac{{n-1 \choose r-1}}{{n \choose r}} = \frac{(n-1)!}{(r-1)!(n-r)!}\div\frac{n!}{r!(n-r)!}= \frac{(n-1)!}{(r-1)!(n-r)!}\cdot\frac{r!(n-r)!}{n!} = \mathbf{\frac{r}{n}}$$

Sampling with replacement

First a clarification. When sampling without replacement, the maximum number of times $x^*$ can appear is, of course, $1$. When sampling with replacement, it can appear between $0$ and $r$ times. Judging by the answer you gave, the question you want to answer is the number of ways the fixed element $x^*$ appears at least once. That is most easily addressed by realizing it is all possible ways except for the times that it does not appear at all. In other words: $$P(\textrm{at least once}) = 1 - P(\textrm{never})$$ In order for $x^*$ to never appear, it cannot appear in any of the $r$ slots, which means that we can only pick from the remaining $n-1$ items. The probability that in slot $1$ we select something other than $x^*$ is $\frac{n-1}{n} = 1 - \frac{1}{n}$. This has to happen in every one of the $r$ slots, so the probability of having no manifestations of the fixed element in a sample of size $r$ is: $$\left(\frac{n-1}{n}\right)^r = \left(1 - \frac{1}{n}\right)^r$$ So the probability of at least one showing is everything else, or: $$\mathbf{1- \left(1 - \frac{1}{n}\right)^r}$$

Unfortunately, this is not the same as the value you posted in the question. Is it at all possible that the value in the innermost parentheses was supposed to be $1-\frac{1}{n}$ and not $\frac{1}{1-n}$?

for without replacement :- total no. of possible ways of selecting $r$ elements from $n$ elements = $_nC_r$

total no. of ways where element $x$ is always selected would be equal to selecting $(r-1)$ element from $(n-1)$ elements [as we would consider $x$ to be already selected] ,

which would be = $_{(n-1)}C_{(r-1)}$

probability = $$\frac{{_{(n-1)}}C_{(r-1)}}{_nC_r} = \frac{r}{n}$$

:-> $C$ is the combination

for with replacement :-

total possible no. of selections would be = C(n+r-1,r) [bars and star logic]

total cases where element x is never selected are = C(n+r-2,r) [n reduces to n-1]

probability of at least one selection of element x = [1 - {C(n+r-2,r)}/{C(n+r-1,r)}]

which comes out to be = (n-1)/(n+r-1)