Probability of min(x,y)<1 [on hold]












-1














How can I solve this?



$f_{xy}(x,y)$ is a joint probability distribution defined by



$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And for other $x,y$ the joint probability distribution defiend by
$$f_{xy}(x,y)=0$$



And the Question is Find the probability of:
P(min(x,y)<1)=?
Please Write Your Full Answer.










share|cite|improve this question







New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
    – angryavian
    yesterday






  • 1




    Possible duplicate of Can You Help me ?A simple Probability Question
    – StubbornAtom
    yesterday
















-1














How can I solve this?



$f_{xy}(x,y)$ is a joint probability distribution defined by



$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And for other $x,y$ the joint probability distribution defiend by
$$f_{xy}(x,y)=0$$



And the Question is Find the probability of:
P(min(x,y)<1)=?
Please Write Your Full Answer.










share|cite|improve this question







New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
    – angryavian
    yesterday






  • 1




    Possible duplicate of Can You Help me ?A simple Probability Question
    – StubbornAtom
    yesterday














-1












-1








-1







How can I solve this?



$f_{xy}(x,y)$ is a joint probability distribution defined by



$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And for other $x,y$ the joint probability distribution defiend by
$$f_{xy}(x,y)=0$$



And the Question is Find the probability of:
P(min(x,y)<1)=?
Please Write Your Full Answer.










share|cite|improve this question







New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











How can I solve this?



$f_{xy}(x,y)$ is a joint probability distribution defined by



$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And for other $x,y$ the joint probability distribution defiend by
$$f_{xy}(x,y)=0$$



And the Question is Find the probability of:
P(min(x,y)<1)=?
Please Write Your Full Answer.







probability






share|cite|improve this question







New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Mobina K

134




134




New contributor




Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac yesterday


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
    – angryavian
    yesterday






  • 1




    Possible duplicate of Can You Help me ?A simple Probability Question
    – StubbornAtom
    yesterday














  • 1




    Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
    – angryavian
    yesterday






  • 1




    Possible duplicate of Can You Help me ?A simple Probability Question
    – StubbornAtom
    yesterday








1




1




Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
– angryavian
yesterday




Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
– angryavian
yesterday




1




1




Possible duplicate of Can You Help me ?A simple Probability Question
– StubbornAtom
yesterday




Possible duplicate of Can You Help me ?A simple Probability Question
– StubbornAtom
yesterday










1 Answer
1






active

oldest

votes


















0














$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    $$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
    $$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
    So we have that
    $$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$






    share|cite|improve this answer


























      0














      $$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
      $$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
      So we have that
      $$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$






      share|cite|improve this answer
























        0












        0








        0






        $$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
        $$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
        So we have that
        $$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$






        share|cite|improve this answer












        $$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
        $$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
        So we have that
        $$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Hello_World

        4,11621630




        4,11621630















            Popular posts from this blog

            Mario Kart Wii

            What does “Dominus providebit” mean?

            Antonio Litta Visconti Arese