Inclusion-Exclusion Principle/Examples/3 Events in Event Space: Example

From ProofWiki
Jump to navigation Jump to search

Examples of Use of Inclusion-Exclusion Principle

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $A, B, C \in \Sigma$ such that:.

Then:

\(\ds \map \Pr A\) \(=\) \(\ds \dfrac 5 {10}\)
\(\ds \map \Pr B\) \(=\) \(\ds \dfrac 7 {10}\)
\(\ds \map \Pr C\) \(=\) \(\ds \dfrac 6 {10}\)
\(\ds \map \Pr {A \cap B}\) \(=\) \(\ds \dfrac 3 {10}\)
\(\ds \map \Pr {B \cap C}\) \(=\) \(\ds \dfrac 4 {10}\)
\(\ds \map \Pr {A \cap C}\) \(=\) \(\ds \dfrac 2 {10}\)
\(\ds \map \Pr {A \cap B \cap C}\) \(=\) \(\ds \dfrac 1 {10}\)

The probability that exactly $2$ of the events $A$, $B$ and $C$ occur is $\dfrac 6 {10}$.


Proof

We are looking for the probability of:

$\paren {\paren {A \cap B} \setminus \paren {A \cap B \cap C} } \cup \paren {\paren {B \cap C} \setminus \paren {A \cap B \cap C} } \cup \paren {\paren {A \cap C} \setminus\paren {A \cap B \cap C} }$

We have that:

$\paren {\paren {A \cap B} \setminus \paren {A \cap B \cap C} } \cap \paren {A \cap B \cap C} = \O$

and similarly for the other two such terms.

Then we have that $\paren {A \cap B} \setminus \paren {A \cap B \cap C}$, $\paren {\paren {B \cap C} \setminus \paren {A \cap B \cap C} }$ and $\paren {\paren {A \cap C} \setminus \paren {A \cap B \cap C} }$ are pairwise disjoint.

Hence the probability $P$ that exactly $2$ of the events $A$, $B$ and $C$ occur is

\(\ds P\) \(=\) \(\ds \paren {\map \Pr {A \cap B} - \map \Pr {A \cap B \cap C} } - \paren {\map \Pr {B \cap C} - \map \Pr {A \cap B \cap C} } - \paren {\map \Pr {A \cap C} - \map \Pr {A \cap B \cap C} }\)
\(\ds \) \(=\) \(\ds \paren {\dfrac 3 {10} - \dfrac 1 {10} } - \paren {\dfrac 4 {10} - \dfrac 1 {10} } - \paren {\dfrac 2 {10} - \dfrac 1 {10} }\)
\(\ds \) \(=\) \(\ds \dfrac 6 {10}\)

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Inclusion-Exclusion_Principle/Examples/3_Events_in_Event_Space:_Example&oldid=496262"