Practice Problems

Categories
Unit 01 Unit 02 Unit 03 Unit 04 Unit 05 Unit 06 Unit 07
Edit Problem
Unit 01 /
Problem 2:


E using axioms to prove min or max
|
Easy


Prove Bonferroni's inequality:
$$P(A\cap B)\geq P(A) + P(B) - 1.$$ Hint: The addition rule and Kolmogorov's axioms.

Solution:

\[\begin{align}\text{Restate & Bonferroni's inequality}\\
P(A\cap B) &\geq P(A) + P(B) - 1\\
\text{Addition Rule &Thrm 1.1.7}\\
P(A\cup B) &= P(A) + P(B) - P(A\cap B)\\
0 &= P(A) + P(B) - P(A\cap B) - P(A\cup B) \\
P(A\cap B) &= P(A) + P(B) - P(A\cup B)\\
\text{Plug in to & Bonferroni's inequality}\\
P(A) + P(B) - P(A\cup B) &\geq P(A) + P(B) - 1\\
- P(A\cup B) &\geq (-1) \\
P(A\cup B) &\leq 1\\
\end{align}\]

From Thrm 1.1.4 basic properties of probability (probability bounds),we know that the statement $P(A\cup B) \leq 1$ is true. Therefore thesupposition that led to it is also true. We have proven Bonferroni'sinequality.

(For those keeping score, the Kolgomorov axioms (Defn 1.1.2) come intothis because they are the direct source of our basic properties ofprobability presented in Thrm 1.1.4)