Can't find the distribution for the random two-dimensional variable
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So I have an assignment regarding probability that states the following:
Six terminals are connected in a communication line. Exactly 3 of the 6 terminals are functional and 3 are not. Each terminal has a fair chance of being picked of if we define the event A which states that we pick a functional terminal from the 6, then P(A)=0.5.
The terminals are being inspected in order until we find a functional terminal.
Let X be the random variable that defines the number of inspections we need to make until we find the first functional terminal, and Y be the random variable that defines the number of inspections we need to make until we find the second working terminal.
We need to find the law of distribution for the two-dimensional random variable {X,Y}.
I've managed to find the marginal distribution of both X and Y separately but I have no idea how to the distribution for the vector {X,Y}.
And ideas?
probability random-variables
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
$endgroup$
add a comment |
$begingroup$
So I have an assignment regarding probability that states the following:
Six terminals are connected in a communication line. Exactly 3 of the 6 terminals are functional and 3 are not. Each terminal has a fair chance of being picked of if we define the event A which states that we pick a functional terminal from the 6, then P(A)=0.5.
The terminals are being inspected in order until we find a functional terminal.
Let X be the random variable that defines the number of inspections we need to make until we find the first functional terminal, and Y be the random variable that defines the number of inspections we need to make until we find the second working terminal.
We need to find the law of distribution for the two-dimensional random variable {X,Y}.
I've managed to find the marginal distribution of both X and Y separately but I have no idea how to the distribution for the vector {X,Y}.
And ideas?
probability random-variables
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
$endgroup$
$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10
add a comment |
$begingroup$
So I have an assignment regarding probability that states the following:
Six terminals are connected in a communication line. Exactly 3 of the 6 terminals are functional and 3 are not. Each terminal has a fair chance of being picked of if we define the event A which states that we pick a functional terminal from the 6, then P(A)=0.5.
The terminals are being inspected in order until we find a functional terminal.
Let X be the random variable that defines the number of inspections we need to make until we find the first functional terminal, and Y be the random variable that defines the number of inspections we need to make until we find the second working terminal.
We need to find the law of distribution for the two-dimensional random variable {X,Y}.
I've managed to find the marginal distribution of both X and Y separately but I have no idea how to the distribution for the vector {X,Y}.
And ideas?
probability random-variables
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
$endgroup$
So I have an assignment regarding probability that states the following:
Six terminals are connected in a communication line. Exactly 3 of the 6 terminals are functional and 3 are not. Each terminal has a fair chance of being picked of if we define the event A which states that we pick a functional terminal from the 6, then P(A)=0.5.
The terminals are being inspected in order until we find a functional terminal.
Let X be the random variable that defines the number of inspections we need to make until we find the first functional terminal, and Y be the random variable that defines the number of inspections we need to make until we find the second working terminal.
We need to find the law of distribution for the two-dimensional random variable {X,Y}.
I've managed to find the marginal distribution of both X and Y separately but I have no idea how to the distribution for the vector {X,Y}.
And ideas?
probability random-variables
probability random-variables
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
asked Dec 16 '18 at 16:46
David DanielsDavid Daniels
132
132
$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10
add a comment |
$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10
$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10
add a comment |
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$begingroup$
In this case, there aren't many possibilities of what the random variable $(X,Y)$ can be. So you can just go through these possibilities one by one. By the way, you can use Latex formatting on this site to make your question look nicer.
$endgroup$
– Tki Deneb
Dec 16 '18 at 17:00
$begingroup$
Thanks for the hint. Looks like I understood how the process works. If say we are looking for P{X=1, Y=3} then we need to find the probability of finding a working terminal in the first and third inspection: P{X=1, Y=3}=P(A)*P(B)*P(A), (P(B) is the probability that will find a terminal that isn't working) therefore P{X=1, Y=3}=3/6*3/5*2/4.
$endgroup$
– David Daniels
Dec 16 '18 at 22:10