wykazac nierownosc

[annoo]

Wykaz, ze jezeli

\(\displaystyle{ A,B \subset \Omega}\)
\(\displaystyle{ P(A)= \frac{1}{4}}\)
\(\displaystyle{ P(B)= \frac{1}{3}}\)

to:

\(\displaystyle{ \frac{1}{3} \le P(A \cup B) \le \frac{7}{12}}\)
\(\displaystyle{ P(B-A) \ge \frac{1}{12}}\)

[JakimPL]

\(\displaystyle{ P(A \cup B) = P(A) + P(B) - P(A \cap B)}\)

\(\displaystyle{ P(A \cup B) = \frac{1}{4} + \frac{1}{3} - P(A \cap B)}\)

\(\displaystyle{ P(A \cup B) = \frac{7}{12} - P(A \cap B)}\)

I teraz, najbardziej skrajne możliwości to takie:
\(\displaystyle{ 1. \ A \subset B \Rightarrow P(A \cap B)=P(A)=\frac{1}{4} \Rightarrow P(A \cup B) = \frac{7}{12} - \frac{1}{4} = \frac{4}{12} = \frac{1}{3}}\)
\(\displaystyle{ 2. \ A \cap B = \emptyset \Rightarrow P(A \cap B) =0 \Rightarrow P(A \cup B) = \frac{7}{12}}\)

Na tej podstawie wynika \(\displaystyle{ \frac{1}{3} \le P(A \cup B) \le \frac{7}{12}}\).