91Ó°ÊÓ

A box contain red and blue balls.

N be the no. of balls in the box.

X be the no. of red balls in the sample

Proportion of the red balls is p, hence there are np red balls in the box.

There are\(N\left( {1 - p} \right)\)blue balls in the box.

Let\(k = Np\)hence there are N-K blue balls and K red balls.

If n>K ,then\({\rm P}\left( {Y = n} \right) = 0\)

Since there are not enough red balls.

Since\({\rm P}\left( {X = n} \right) > 0\)foe all n , the result is true if n>K.

For\(n \le K\)Let\({X_i} = 1\)if ith ball is red for i=1,...n.for sampling without replacement.

\(\begin{aligned}{c}{\rm P}\left( {X = n} \right) = {\rm P}\left( {{X_1} = 1} \right)\\ = \prod\limits_{}^n {{\rm P}\left( {{X_i} = 1|{X_1} = 1,...{X_{i - 1}} = 1} \right)} \\ = \frac{K}{N}\frac{{K - 1}}{{N - 1}}...\frac{{K - n + 1}}{{N - n + 1}}\end{aligned}\)......(1)

For sampling with replacement the\({X_i}\)are independent, therefore

\(\begin{aligned}{c}{\rm P}\left( {Y = n} \right) = \prod\limits_{i = 1}^n {{\rm P}\left( {{Y_i} = 1} \right)} \\ = {\left( {\frac{K}{N}} \right)^n}\end{aligned}\)..............(2)

For\(j = 1...,n - 1,KN - jn < KN - jK\)so\(\frac{{\left( {K - j} \right)}}{{\left( {N - j} \right)}} < \frac{K}{N}\).

Product in (1)is smaller than (2), only if N is finite.

That is.\({\rm P}\left( {X = n} \right) > {\rm P}\left( {Y = n} \right)\)

If N is infinite, then sampling with and without replacement equivalent.