Probability of min(x,y)<1 [closed]
$begingroup$
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
asked Jan 5 at 6:14
Mobina KMobina K
194
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closed as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac Jan 5 at 9:06
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.
add a comment |
$begingroup$
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
asked Jan 5 at 6:14
Mobina KMobina K
194
$endgroup$
closed as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac Jan 5 at 9:06
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
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
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– angryavian
Jan 5 at 6:18
1
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27
add a comment |
$begingroup$
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
asked Jan 5 at 6:14
Mobina KMobina K
194
$endgroup$
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
probability
asked Jan 5 at 6:14
Mobina KMobina K
194
asked Jan 5 at 6:14
Mobina KMobina K
194
asked Jan 5 at 6:14
Mobina KMobina K
194
asked Jan 5 at 6:14
Mobina KMobina K
194
asked Jan 5 at 6:14
Mobina KMobina K
194
194
closed as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac Jan 5 at 9:06
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.
closed as off-topic by Eevee Trainer, StubbornAtom, Lee David Chung Lin, Saad, caverac Jan 5 at 9:06
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
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
$endgroup$
– angryavian
Jan 5 at 6:18
1
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27
add a comment |
1
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
$endgroup$
– angryavian
Jan 5 at 6:18
1
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27
1
1
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
$endgroup$
– angryavian
Jan 5 at 6:18
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
$endgroup$
– angryavian
Jan 5 at 6:18
1
1
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27
add a comment |
1 Answer
1
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oldest
votes
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$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
$endgroup$
add a comment |
$begingroup$
$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
$endgroup$
add a comment |
$begingroup$
$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
$endgroup$
$$P(min(x,y)<1)=1-P(min(x,y)geq 1) = 1-P(xgeq 1,ygeq 1)$$
$$ = 1-int_{1}^{infty} int_{1}^{infty} ye^{-y(x+1)}dxdy.$$
So we have that
$$P(min(x,y)<1) = 1-int_{1}^{infty}e^{-(y+1)}dy=1-frac{1}{e^2}.$$
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
answered Jan 5 at 6:21
model_checkermodel_checker
4,14621931
4,14621931
add a comment |
add a comment |
1
$begingroup$
Draw the region ${min(x,y) < 1}$ on the $x$-$y$ plane and integrate the density over that region.
$endgroup$
– angryavian
Jan 5 at 6:18
1
$begingroup$
Possible duplicate of Can You Help me ?A simple Probability Question
$endgroup$
– StubbornAtom
Jan 5 at 7:27