Find the joint distribution with 2 discrete random variables
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2
down vote
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How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:
P(X=0, Y=0) = (0.4)(0.8) = 0.32
P(X=1, Y=1) = (1-0.4)(0.9) = 0.54
How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?
probability probability-theory probability-distributions
add a comment |
up vote
2
down vote
favorite
How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:
P(X=0, Y=0) = (0.4)(0.8) = 0.32
P(X=1, Y=1) = (1-0.4)(0.9) = 0.54
How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?
probability probability-theory probability-distributions
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:
P(X=0, Y=0) = (0.4)(0.8) = 0.32
P(X=1, Y=1) = (1-0.4)(0.9) = 0.54
How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?
probability probability-theory probability-distributions
How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:
P(X=0, Y=0) = (0.4)(0.8) = 0.32
P(X=1, Y=1) = (1-0.4)(0.9) = 0.54
How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked 2 days ago
Yolanda Hui
13710
13710
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add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint:
$P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:
$P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
|
show 1 more comment
up vote
1
down vote
accepted
Hint:
$P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
|
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:
$P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$
Hint:
$P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$
answered 2 days ago
drhab
94.6k543125
94.6k543125
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
|
show 1 more comment
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
– Yolanda Hui
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
– drhab
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
– Yolanda Hui
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
$P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
– drhab
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
– Yolanda Hui
2 days ago
|
show 1 more comment
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