When do i use the addition law of probability??? [closed]

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Hi im really confused on when and how to use the addition law of probability. Does this law apply to all P(A or B)??? because I used when it is asking the probability of getting a queen or king but it doesnt work. Please help thanks!






probability







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closed as off-topic by anomaly, Math1000, Delta-u, user91500, Hans Lundmark Sep 3 at 10:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anomaly, Math1000, Delta-u, user91500, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
    – HJ_beginner
    Sep 3 at 7:44










  • Thanks! God bless you only you willing to help me!
    – MERcurialKG
    Sep 3 at 8:06






  • 1




    Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
    – HJ_beginner
    Sep 3 at 8:09










  • well said @HJ_beginner
    – MERcurialKG
    Sep 3 at 8:29














up vote
-5
down vote

favorite












Hi im really confused on when and how to use the addition law of probability. Does this law apply to all P(A or B)??? because I used when it is asking the probability of getting a queen or king but it doesnt work. Please help thanks!






probability







share|cite|improve this question













closed as off-topic by anomaly, Math1000, Delta-u, user91500, Hans Lundmark Sep 3 at 10:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anomaly, Math1000, Delta-u, user91500, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
    – HJ_beginner
    Sep 3 at 7:44










  • Thanks! God bless you only you willing to help me!
    – MERcurialKG
    Sep 3 at 8:06






  • 1




    Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
    – HJ_beginner
    Sep 3 at 8:09










  • well said @HJ_beginner
    – MERcurialKG
    Sep 3 at 8:29












up vote
-5
down vote

favorite









up vote
-5
down vote

favorite











Hi im really confused on when and how to use the addition law of probability. Does this law apply to all P(A or B)??? because I used when it is asking the probability of getting a queen or king but it doesnt work. Please help thanks!






probability







share|cite|improve this question













Hi im really confused on when and how to use the addition law of probability. Does this law apply to all P(A or B)??? because I used when it is asking the probability of getting a queen or king but it doesnt work. Please help thanks!






probability




probability






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share|cite|improve this question




share|cite|improve this question










asked Sep 3 at 7:12









MERcurialKG

146




146




closed as off-topic by anomaly, Math1000, Delta-u, user91500, Hans Lundmark Sep 3 at 10:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anomaly, Math1000, Delta-u, user91500, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by anomaly, Math1000, Delta-u, user91500, Hans Lundmark Sep 3 at 10:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anomaly, Math1000, Delta-u, user91500, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    $P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
    – HJ_beginner
    Sep 3 at 7:44










  • Thanks! God bless you only you willing to help me!
    – MERcurialKG
    Sep 3 at 8:06






  • 1




    Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
    – HJ_beginner
    Sep 3 at 8:09










  • well said @HJ_beginner
    – MERcurialKG
    Sep 3 at 8:29












  • 2




    $P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
    – HJ_beginner
    Sep 3 at 7:44










  • Thanks! God bless you only you willing to help me!
    – MERcurialKG
    Sep 3 at 8:06






  • 1




    Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
    – HJ_beginner
    Sep 3 at 8:09










  • well said @HJ_beginner
    – MERcurialKG
    Sep 3 at 8:29







2




2




$P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
– HJ_beginner
Sep 3 at 7:44




$P(A cup B) = P(A) + P(B)$ works if $A$ and $B$ are mutually exclusive. So if you draw one card from a random deck, then $P(Queen cup King) = P(Queen) + P(King)$ because if you draw only one card then you don't have to worry about getting two cards on a single draw (ie a Queen and King at the same time which is impossible).
– HJ_beginner
Sep 3 at 7:44












Thanks! God bless you only you willing to help me!
– MERcurialKG
Sep 3 at 8:06




Thanks! God bless you only you willing to help me!
– MERcurialKG
Sep 3 at 8:06




1




1




Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
– HJ_beginner
Sep 3 at 8:09




Hey no problem brother we're all citizens of the universe. Nobody knows what we're doing here or what this is all about, all we have is each other.
– HJ_beginner
Sep 3 at 8:09












well said @HJ_beginner
– MERcurialKG
Sep 3 at 8:29




well said @HJ_beginner
– MERcurialKG
Sep 3 at 8:29










1 Answer
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In general $$P(Acup B)=P(A)+P(B)-P(Acap B)$$



In special case that $Acap B=varnothing$ (i.e. the events are mutually exclusive) this becomes:$$P(Acup B)=P(A)+P(B)$$ on base of $P(varnothing)=0$.



Here $cup$ must be red as "or" and $cap$ as "and".



P.S. also ask God to bless me, okay? :-)






share|cite|improve this answer



























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    In general $$P(Acup B)=P(A)+P(B)-P(Acap B)$$



    In special case that $Acap B=varnothing$ (i.e. the events are mutually exclusive) this becomes:$$P(Acup B)=P(A)+P(B)$$ on base of $P(varnothing)=0$.



    Here $cup$ must be red as "or" and $cap$ as "and".



    P.S. also ask God to bless me, okay? :-)






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      In general $$P(Acup B)=P(A)+P(B)-P(Acap B)$$



      In special case that $Acap B=varnothing$ (i.e. the events are mutually exclusive) this becomes:$$P(Acup B)=P(A)+P(B)$$ on base of $P(varnothing)=0$.



      Here $cup$ must be red as "or" and $cap$ as "and".



      P.S. also ask God to bless me, okay? :-)






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        In general $$P(Acup B)=P(A)+P(B)-P(Acap B)$$



        In special case that $Acap B=varnothing$ (i.e. the events are mutually exclusive) this becomes:$$P(Acup B)=P(A)+P(B)$$ on base of $P(varnothing)=0$.



        Here $cup$ must be red as "or" and $cap$ as "and".



        P.S. also ask God to bless me, okay? :-)






        share|cite|improve this answer












        In general $$P(Acup B)=P(A)+P(B)-P(Acap B)$$



        In special case that $Acap B=varnothing$ (i.e. the events are mutually exclusive) this becomes:$$P(Acup B)=P(A)+P(B)$$ on base of $P(varnothing)=0$.



        Here $cup$ must be red as "or" and $cap$ as "and".



        P.S. also ask God to bless me, okay? :-)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 3 at 8:09









        drhab

        89k541122




        89k541122












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