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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, $P(A ∩ B) ≤ P(A ∪ B) ≤ P(A) + P(B)$

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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