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, $P(A ∩ B) ≤ P(A ∪ B) ≤ P(A) + P(B)$

  • $\begingroup$ What does $P(X)$ mean? The power set of $X$? The probability of an event $X$? What does P(X) \le P(Y)$ mean? $\endgroup$ – fleablood Jan 26 at 17:51
  • 2
    $\begingroup$ @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. $\endgroup$ – JMoravitz Jan 26 at 18:08
  • $\begingroup$ If you use Venn diagram the problem will simplify and you will be able to visualize better $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.