Proving that if $A$ and $B$ are events, then $Pr[A ∩ B] ge Pr[A] + Pr[B] - 1$Probability of the union of $3$ events?$K$ events that are $(K-1)$-wise Independent but not Mutually/Fully IndependentProbability of a Union of EventsProving that three events are mutually independentProving that $Pleft ( bigcup_i=1^nA_i right )leq sum_i=1^nP(A_i)$ by inductionProbability: Prove that events are independentCan two different events generate the same set of probabilities?If $A$ and B are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$Given $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Bounding the probability of a set of non independent events

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Proving that if $A$ and $B$ are events, then $Pr[A ∩ B] ge Pr[A] + Pr[B] - 1$


Probability of the union of $3$ events?$K$ events that are $(K-1)$-wise Independent but not Mutually/Fully IndependentProbability of a Union of EventsProving that three events are mutually independentProving that $Pleft ( bigcup_i=1^nA_i right )leq sum_i=1^nP(A_i)$ by inductionProbability: Prove that events are independentCan two different events generate the same set of probabilities?If $A$ and B are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$Given $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Bounding the probability of a set of non independent events













1












$begingroup$


I am just looking for some guidance whether I am approaching the questions correctly. New to probability!




Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$




I rewrote the statement in the following way :



$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$



And based on some probability properties, I rewrote out the statement again:



$$1 geq Pr[A cup B]$$



Is this enough to prove the statement?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Perfect! ... ...
    $endgroup$
    – Maria Mazur
    Mar 20 at 17:51















1












$begingroup$


I am just looking for some guidance whether I am approaching the questions correctly. New to probability!




Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$




I rewrote the statement in the following way :



$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$



And based on some probability properties, I rewrote out the statement again:



$$1 geq Pr[A cup B]$$



Is this enough to prove the statement?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Perfect! ... ...
    $endgroup$
    – Maria Mazur
    Mar 20 at 17:51













1












1








1





$begingroup$


I am just looking for some guidance whether I am approaching the questions correctly. New to probability!




Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$




I rewrote the statement in the following way :



$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$



And based on some probability properties, I rewrote out the statement again:



$$1 geq Pr[A cup B]$$



Is this enough to prove the statement?










share|cite|improve this question











$endgroup$




I am just looking for some guidance whether I am approaching the questions correctly. New to probability!




Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$




I rewrote the statement in the following way :



$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$



And based on some probability properties, I rewrote out the statement again:



$$1 geq Pr[A cup B]$$



Is this enough to prove the statement?







probability proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 19:55









Maria Mazur

49.5k1361124




49.5k1361124










asked Mar 20 at 17:49









RobinRobin

525




525







  • 2




    $begingroup$
    Perfect! ... ...
    $endgroup$
    – Maria Mazur
    Mar 20 at 17:51












  • 2




    $begingroup$
    Perfect! ... ...
    $endgroup$
    – Maria Mazur
    Mar 20 at 17:51







2




2




$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51




$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$



So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.



That is correct.






share|cite|improve this answer









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    1












    $begingroup$

    So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$



    So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.



    That is correct.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$



      So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.



      That is correct.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$



        So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.



        That is correct.






        share|cite|improve this answer









        $endgroup$



        So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$



        So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.



        That is correct.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 20 at 17:52









        Maria MazurMaria Mazur

        49.5k1361124




        49.5k1361124



























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