91Ó°ÊÓ

Consider $n$independent trials, each of which results in one of the outcomes $1,...,k$ with respective probabilities$p_1,…,p_k,\underset{i=1}{\overset{k}{∑}}{p}_i=1$.

For the opening, consider only two trials. Because of the independence, the probability that in these two trials we will get the exact results is simply

$p=\underset{j=1}{\overset{k}{∑}}{p}_j^2$

Now, let's look at all possible pairs of $n$trials. There exist $\left(\begin{array}{l}n\\ 2\end{array}\right)=\frac{n(n-1)}{2}$such combinations.

The probability that each pair has the property that they have the same outcome is $p$.

Define $X$as the random variable that marks the number of pairs that have the same outcome. Use the Poisson property to obtain that $X$has approximately $Pois\left(\mathrm{λ}\right)$distribution, where

$\mathrm{λ}=\left(\begin{array}{l}n\\ 2\end{array}\right)×p=\frac{n(n-1)}{2}·\underset{j=1}{\overset{k}{∑}}{p}_j^2$

Hence, the probability that there will not be such a pair is

$P(X=0)=\exp (-\mathrm{λ})=\exp \left(-\frac{n(n-1)}{2}·\underset{j=1}{\overset{k}{∑}}{p}_j^2\right)$

Which had to be shown.

The probability that there will not be such a pair is $P(X=0)=\exp (-\mathrm{λ})=\exp \left(-\frac{n(n-1)}{2}·\underset{j=1}{\overset{k}{∑}}{p}_j^2\right)$