# Prove that $P(A ∩ B) ≤ P(A ∪ B) ≤ P(A) + P(B)$

I don´t know if my proof is right. I separated it into 2:

$$1) P(A ∩ B) ≤ P(A ∪ B)$$

Pf: $$P(A ∪ B) = P(A\setminus B) + P(B\setminus A) + P(A ∩ B)$$ and we know that $$0 ≤ P(A\setminus B) ≤ 1$$ and $$0 ≤ P(B\setminus A) ≤ 1$$ so $$P(A ∩ B) ≤ P(A ∪ B)$$

$$2) P(A ∪ B) ≤ P(A) + P(B)$$

Pf: $$P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$$ and we know that $$0 ≤ P(A ∩ B) ≤ 1$$ so $$P(A ∪ B) ≤ P(A) + P(B)$$

Therefore, $$P(A ∩ B) ≤ P(A ∪ B) ≤ P(A) + P(B)$$

• What does $P(X)$ mean? The power set of $X$? The probability of an event $X$? What does P(X) \le P(Y)$mean? – fleablood Jan 26 at 17:51 • @fleablood from context,$P(X)$is clearly referring to probability and$\le$means the usual less than or equal relation on real numbers. @ Regina, you have some misused symbols. Set difference is \ , not to be confused with / which is more often for division or quotient groups etc... Visit this page for information on how to type with MathJax and$\LaTeX\$ on this site. – JMoravitz Jan 26 at 18:08
• If you use Venn diagram the problem will simplify and you will be able to visualize better – SNEHIL SANYAL Jan 27 at 4:08

Looks fine. In the first proof you are really making use of the fact that $$P(A\setminus B)\geq 0$$ and $$P(B\setminus A)\geq 0$$. In the second proof you are making use of the fact that $$P(A\cap B)\geq 0$$. You aren't really making use of the fact that the probabilities are at most one so I would omit it from the explanation.