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Let\({A_1},{A_2}, \ldots \)be an arbitrary infinite sequence of events.

Let\({B_1},{B_2}, \ldots \)be another infinite sequence of events.

\({B_1} = {A_1}\), \({B_2} = {A_1}^c \cap {A_2}\), \({B_3} = {A_1}^c \cap {A_2}^c \cap {A_3}\), \({B_4} = {A_1}^c \cap {A_2}^c \cap {A_3}^c \cap {A_4}\),鈥�

By using the principle of mathematical induction:

Put i=2,

Therefore,

\(\begin{aligned}{c}\Pr \left( {\bigcup\limits_{i = 1}^2 {{A_i}} } \right) &= \Pr \left( {{A_1} \cup {A_2}} \right)\\ &= \Pr \left( {{A_1}} \right) + \Pr \left( {{A_2}} \right) - \Pr \left( {{A_1} \cap {A_2}} \right)\\ &= \Pr \left( {{B_1}} \right) + \Pr \left( {{A_1}^c \cap {A_2}} \right)\\ &= \Pr \left( {{B_1}} \right) + \Pr \left( {{B_2}} \right)\\ &= \sum\limits_{i = 1}^2 {\Pr \left( {{B_i}} \right)} \end{aligned}\)

That is,

\(\Pr \left( {\bigcup\limits_{i = 1}^2 {{A_i}} } \right) = \sum\limits_{i = 1}^2 {\Pr \left( {{B_i}} \right)} \)

Assume that it is true for n=k.

Therefore,

\(\Pr \left( {\bigcup\limits_{i = 1}^k {{A_i}} } \right) = \sum\limits_{i = 1}^k {\Pr \left( {{B_i}} \right)} \)

We need to check whether it is true for n=k+1.

\(\begin{aligned}{c}\Pr \left( {\bigcup\limits_{i = 1}^{k + 1} {{A_i}} } \right) &= \Pr \left( {\bigcup\limits_{i = 1}^k {{A_i} \cup {A_{k + 1}}} } \right)\\ &= \Pr \left( {\bigcup\limits_{i = 1}^k {{A_i}} } \right) + \Pr \left( {{A_{k + 1}}} \right) - \Pr \left( {\bigcup\limits_{i = 1}^k {{A_i}} \cap {A_{k + 1}}} \right)\\ &= \Pr \left( {\bigcup\limits_{i = 1}^k {{A_i}} } \right) + \Pr \left( {{{\left( {\bigcup\limits_{i = 1}^k {{A_i}} } \right)}^c} \cap {A_{k + 1}}} \right)\\ &= \Pr \left( {{B_k}} \right) + \Pr \left( {{A_1}^c \cap {A_2}^c \cap {A_3}^c \ldots \cap {A_k}^c \cap {A_{k + 1}}} \right)\\ &= \Pr \left( {{B_k}} \right) + \Pr \left( {{B_{k + 1}}} \right)\\ &= \sum\limits_{i = 1}^{k + 1} {\Pr \left( {{B_{k + 1}}} \right)} \end{aligned}\)

That is:

\(\Pr \left( {\bigcup\limits_{i = 1}^{k + 1} {{A_i}} } \right) = \sum\limits_{i = 1}^{k + 1} {\Pr \left( {{B_i}} \right)} \)

Hence by the principle of mathematical induction:

\(\Pr \left( {\bigcup\limits_{i = 1}^n {{A_i}} } \right) = \sum\limits_{i = 1}^n {\Pr \left( {{B_i}} \right)} \)

\(\begin{aligned}{c}Pr\left( {\bigcup\limits_{i = 1}^\infty {{A_i}} } \right) &= Pr\left( {\bigcup\limits_{i = 1}^n {{A_i}} \bigcup\limits_{k = n + 1}^\infty {{A_k}} } \right)\\ &= Pr\left( {\bigcup\limits_{i = 1}^n {{A_i}} } \right) + Pr\left( {\bigcup\limits_{k = n + 1}^\infty {{A_k}} } \right) - Pr\left( {\bigcup\limits_{i = 1}^n {{A_i}} \cap \bigcup\limits_{k = n + 1}^\infty {{A_k}} } \right)\\ &= \sum\limits_{i = 1}^n {\Pr \left( {{{\mathop{\rm B}\nolimits} _i}} \right)} + Pr\left( {{{\left( {\bigcup\limits_{i = 1}^n {{A_i}} } \right)}^c} \cap \bigcup\limits_{k = n + 1}^\infty {{A_k}} } \right)\\ &= \sum\limits_{i = 1}^n {\Pr \left( {{{\mathop{\rm B}\nolimits} _i}} \right)} + Pr\left( {\bigcup\limits_{k = n + 1}^\infty {{A_k}} } \right)\\ &= \sum\limits_{i = 1}^n {\Pr \left( {{{\mathop{\rm B}\nolimits} _i}} \right)} + \sum\limits_{k = n + 1}^\infty {\Pr \left( {{{\mathop{\rm B}\nolimits} _k}} \right)} \\ &= \sum\limits_{i = 1}^\infty {\Pr \left( {{{\mathop{\rm B}\nolimits} _i}} \right)} \end{aligned}\)

Hence,

\(\Pr \left( {\bigcup\limits_{i = 1}^\infty {{A_i}} } \right) = \sum\limits_{i = 1}^\infty {\Pr \left( {{B_i}} \right)} \)