P(A)=\frac{n(A)}{n}

**dwsmith** · January 25th 2011, 11:55 AM

Suppose that an experiment is performed n times. For any event A of the experiment, let n(A) denote the number of times that event A occurs. The relatively frequency definition of probability would propose that

$\displaystyle P(A)=\frac{n(A)}{n}$ .

Prove that this def. satisfies the three axioms of prob.

1 and 2 $0\leq P(A)\leq P(S)=1\Rightarrow 0\leq P(A)\leq 1$

$\displaystyle P(A)\in [0,1]=\frac{n(A)}{n}\Rightarrow n\in [0,1]=n(A)\Rightarrow 0\leq n(A)\leq n=1$

3 $A_i\cap A_j=\emptyset, \ \ i\neq j$ , then $\displaystyle P\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1} ^{\infty}P(A_i)$

Not sure what to do with 3.

**theodds** · January 25th 2011, 12:44 PM

For disjoint sets $A_1, ..., A_n$ it happens to be the case that, using your notation,

$<br /> n\left(A_1 \cup \cdots \cup A_k\right) = \sum_{i = 1} ^ k n(A_i).<br />$

Note that because n < infinity, you can have at most n non-null disjoint events so it suffices to show finite additivity as opposed to countably infinite.

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