شاه کلید نکات
شاه کلید سوال
اگر S فضای نمونه ای متناهی و ناتهی برای یک آزمایش تصادفی باشد و A و B پیشامدهایی در این فضا باشند، در این صورت:
\(I)\,0 \le P(A) \le 1\)
زیرا : \(A \subseteq S\;\; \Rightarrow \;\;0 \le n\left( A \right) \le .....\; \Rightarrow \;\)
\(\;\frac{0}{{n\left( S \right)}} \le \frac{{n\left( A \right)}}{{n\left( S \right)}} \le .....\; \Rightarrow \;0 \le ..... \le 1\)
\(II)\,P\left( \emptyset \right) = 0\;\;,\;\;P\left( S \right) = 1\)
زیرا : \(P\left( \emptyset \right) = \frac{{n\left( \emptyset \right)}}{{.....}} = \frac{{\;.....\;}}{{.....}} = 0\)
\(\begin{array}{l},\\P\left( S \right) = \frac{{\;.....\;}}{{.....}} = 1\\\\III)\,P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\end{array}\)
زیرا : \(\begin{array}{l}n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\;\;\mathop \Rightarrow \limits^{ \div n\left( S \right)} \;\\\end{array}\)
\(\begin{array}{l}\frac{{n\left( {A \cup B} \right)}}{{n\left( S \right)}} = ........\; + \;........\; = \;........\\\\ \Rightarrow \;P\left( {A \cap B} \right) = ........\; + \;........\; - \;P\left( {A \cap B} \right)\end{array}\)
\(I)\,0 \le P(A) \le 1\)
زیرا : \(A \subseteq S\;\; \Rightarrow \;\;0 \le n\left( A \right) \le n\left( S \right)\; \Rightarrow \;\)
\(\frac{0}{{n\left( S \right)}} \le \frac{{n\left( A \right)}}{{n\left( S \right)}} \le \frac{{n\left( S \right)}}{{n\left( S \right)}}\; \Rightarrow \;0 \le P\left( A \right) \le 1\)
\(II)\,P\left( \emptyset \right) = 0\;\;,\;\;P\left( S \right) = 1\)
زیرا : \(P\left( \emptyset \right) = \frac{{n\left( \emptyset \right)}}{{n\left( S \right)}} = \frac{0}{{n\left( S \right)}} = 0\)
\(\begin{array}{l},\\P\left( S \right) = \frac{{n\left( S \right)}}{{n\left( S \right)}} = 1\\\\III)\,P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\end{array}\)
زیرا : \(n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\;\;\mathop \Rightarrow \limits^{ \div n\left( S \right)} \;\)
\(\begin{array}{l}\frac{{n\left( {A \cup B} \right)}}{{n\left( S \right)}} = \frac{{n\left( A \right)}}{{n\left( S \right)}} + \frac{{n\left( B \right)}}{{n\left( S \right)}} = \frac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}}\\\\ \Rightarrow \;P\left( {A \cup B} \right) = P\left( A \right)\; + \;P\left( B \right)\; - \;P\left( {A \cap B} \right)\end{array}\)
1736019749.png)
