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 $\left[1- \left(\frac{1}{1-n}\right)^r\right]$.

Please help me show how this is proved.

  • Hello, if you can, please explain what it is about this question which makes it so you can't prove it. Do you understand what it's saying, and just don't have any idea where to start, or do you not understand completely what it's saying? Giving more context in your question will help you get more relevant answers and hints. Cheers – 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)

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