Mixing
Compute $P(X + Y = 3)$. [closed]
1
$begingroup$
Let $X ∼ Geo(1/4)$ and $Y ∼ Geo(1/2)$. The random variables $X$ and $Y$ are independent. Compute $P(X + Y = 3)$.
For a geometric distribution $P(X=k)=P(1-p)^{k-1}$ where $k=3$ but which value should I take for $p$? The sum of the two $p$s ($1/4+1/2$)? I tried but it's wrong, can someone help me? (Correct answer is $5/32$)
probability
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
$endgroup$
closed as off-topic by Did, Paul Frost, Cesareo, Kemono Chen, Namaste Jan 24 at 2:46
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." – Did, Paul Frost, Cesareo, Kemono Chen, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
1
$begingroup$
Let $X ∼ Geo(1/4)$ and $Y ∼ Geo(1/2)$. The random variables $X$ and $Y$ are independent. Compute $P(X + Y = 3)$.
For a geometric distribution $P(X=k)=P(1-p)^{k-1}$ where $k=3$ but which value should I take for $p$? The sum of the two $p$s ($1/4+1/2$)? I tried but it's wrong, can someone help me? (Correct answer is $5/32$)
probability
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
$endgroup$
closed as off-topic by Did, Paul Frost, Cesareo, Kemono Chen, Namaste Jan 24 at 2:46
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." – Did, Paul Frost, Cesareo, Kemono Chen, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14
add a comment |
1
1
1
$begingroup$
Let $X ∼ Geo(1/4)$ and $Y ∼ Geo(1/2)$. The random variables $X$ and $Y$ are independent. Compute $P(X + Y = 3)$.
For a geometric distribution $P(X=k)=P(1-p)^{k-1}$ where $k=3$ but which value should I take for $p$? The sum of the two $p$s ($1/4+1/2$)? I tried but it's wrong, can someone help me? (Correct answer is $5/32$)
probability
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
$endgroup$
Let $X ∼ Geo(1/4)$ and $Y ∼ Geo(1/2)$. The random variables $X$ and $Y$ are independent. Compute $P(X + Y = 3)$.
For a geometric distribution $P(X=k)=P(1-p)^{k-1}$ where $k=3$ but which value should I take for $p$? The sum of the two $p$s ($1/4+1/2$)? I tried but it's wrong, can someone help me? (Correct answer is $5/32$)
probability
probability
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
asked Jan 22 at 16:12
Luke MarciLuke Marci
856
856
closed as off-topic by Did, Paul Frost, Cesareo, Kemono Chen, Namaste Jan 24 at 2:46
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." – Did, Paul Frost, Cesareo, Kemono Chen, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Paul Frost, Cesareo, Kemono Chen, Namaste Jan 24 at 2:46
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." – Did, Paul Frost, Cesareo, Kemono Chen, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14
add a comment |
1
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14
1
1
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hints:
- You could calculate $P(X+Y=3)$ $=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$
- Since $X$ and $Y$ are independent $P(X=x,Y=y)=P(X=x)P(Y=y)$
answered Jan 22 at 16:17
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Hints:
- You could calculate $P(X+Y=3)$ $=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$
- Since $X$ and $Y$ are independent $P(X=x,Y=y)=P(X=x)P(Y=y)$
answered Jan 22 at 16:17
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
|
show 2 more comments
2
$begingroup$
Hints:
- You could calculate $P(X+Y=3)$ $=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$
- Since $X$ and $Y$ are independent $P(X=x,Y=y)=P(X=x)P(Y=y)$
answered Jan 22 at 16:17
HenryHenry
101k481168
$endgroup$
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
|
show 2 more comments
2
2
2
$begingroup$
Hints:
- You could calculate $P(X+Y=3)$ $=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$
- Since $X$ and $Y$ are independent $P(X=x,Y=y)=P(X=x)P(Y=y)$
answered Jan 22 at 16:17
HenryHenry
101k481168
$endgroup$
Hints:
- You could calculate $P(X+Y=3)$ $=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$
- Since $X$ and $Y$ are independent $P(X=x,Y=y)=P(X=x)P(Y=y)$
answered Jan 22 at 16:17
HenryHenry
101k481168
answered Jan 22 at 16:17
HenryHenry
101k481168
answered Jan 22 at 16:17
HenryHenry
101k481168
answered Jan 22 at 16:17
HenryHenry
101k481168
101k481168
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
|
show 2 more comments
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
Thank you for the comment but if I compute it, my result is 65/192 and not 5/32. I have $1*9/64+1/2*3/16+1/4*1/4+1/8*1/3$, where I'm wrong?
$endgroup$
– Luke Marci
Jan 22 at 16:36
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
For $P(X=0)$ you seem to be using $1$. Depending on which definition of the geometric distribution you are using, I would have thought it should be either $0$ or $frac14$. Similar issues arise with the other parts of your calculation
$endgroup$
– Henry
Jan 22 at 16:52
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
I'm using $P(X=k)=P(1-p)^{k-1}$ is not correct? Which one should I use?
$endgroup$
– Luke Marci
Jan 22 at 16:54
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Ahh I see to point, my definition is from k=1, the other that you use from k=0, I'll try with the other definition :)
$endgroup$
– Luke Marci
Jan 22 at 16:55
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
$begingroup$
Now it's correct, I just use the definition with $k-1$ at the exponent for $k=1,2,3...$ but I didn't consider the case $P(X=0)$ and $P(Y=0)$. But now that I discover this thing for the geometric distribution I have duct, if I want to calculate $P(X>k)$ it is $(1-p)^{k-1}$ or just $(1-p)^{k}$? (using the definition for geometric distribution described above)
$endgroup$
– Luke Marci
Jan 22 at 17:14
|
show 2 more comments
1
$begingroup$
Hint: the sum of two geometric random variables is not a geometric random variable
$endgroup$
– pwerth
Jan 22 at 16:14