# simplified prove of bonferroni inequality

• January 29th 2010, 11:29 AM
alexandrabel90
simplified prove of bonferroni inequality
If A and B are 2 events, prove that P( A ∩ B ) ≥ 1 - P(A') -P(B') and hence P( A ∩ B ∩ C ) ≥ 1 - P(A') -P(B') - P(C').

thanks! I haven learn about bonferroni inequality..so i dont know how to use it though
• January 29th 2010, 12:01 PM
Plato
Quote:

Originally Posted by alexandrabel90
If A and B are 2 events, prove that P( A ∩ B ) ≥ 1 - P(A') -P(B') and hence P( A ∩ B ∩ C ) ≥ 1 - P(A') -P(B') - P(C').

Note that $P(A\cup B)\le 1$ so $-P(A\cup B)\ge -1$.

$\begin{gathered}
P(A \cap B) = P(A) + P(B) - P(A \cup B) \hfill \\
P(A \cap B) \geqslant 1 - P(A') + 1 - P(B') - 1 \hfill \\
P(A \cap B) \geqslant 1 - P(A') - P(B') \hfill \\
\end{gathered}$