Mixing
How do I myself compute the following probability
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I tried to approximate binomial to a poisson distribution but this did not match the conditions since n is too small.
Afterwards I considered the event Y equal the square root of X as being an intersection of 2 independents events Which means this is equal to the probability of y times the probability of square root of x.
however, here the square root really confuses me and in general I have thought about it too much but I am stuck.
probability
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 16:08
Roy Rizk
887
add a comment |Â
up vote
0
down vote
favorite
I tried to approximate binomial to a poisson distribution but this did not match the conditions since n is too small.
Afterwards I considered the event Y equal the square root of X as being an intersection of 2 independents events Which means this is equal to the probability of y times the probability of square root of x.
however, here the square root really confuses me and in general I have thought about it too much but I am stuck.
probability
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 16:08
Roy Rizk
887
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
In R:i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3))returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulationmean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2))returns 0.066067, correct to three places.
– BruceET
Jul 14 at 16:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I tried to approximate binomial to a poisson distribution but this did not match the conditions since n is too small.
Afterwards I considered the event Y equal the square root of X as being an intersection of 2 independents events Which means this is equal to the probability of y times the probability of square root of x.
however, here the square root really confuses me and in general I have thought about it too much but I am stuck.
probability
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 16:08
Roy Rizk
887
I tried to approximate binomial to a poisson distribution but this did not match the conditions since n is too small.
Afterwards I considered the event Y equal the square root of X as being an intersection of 2 independents events Which means this is equal to the probability of y times the probability of square root of x.
however, here the square root really confuses me and in general I have thought about it too much but I am stuck.
probability
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 16:08
Roy Rizk
887
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
15.7k92047
asked Jul 14 at 16:08
Roy Rizk
887
asked Jul 14 at 16:08
Roy Rizk
887
asked Jul 14 at 16:08
Roy Rizk
887
887
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
In R:i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3))returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulationmean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2))returns 0.066067, correct to three places.
– BruceET
Jul 14 at 16:38
add a comment |Â
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
In R:i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3))returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulationmean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2))returns 0.066067, correct to three places.
– BruceET
Jul 14 at 16:38
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
In R:
i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3)) returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulation mean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2)) returns 0.066067, correct to three places.– BruceET
Jul 14 at 16:38
In R:
i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3)) returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulation mean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2)) returns 0.066067, correct to three places.– BruceET
Jul 14 at 16:38
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
begineqnarray* P(sqrtX=Y) &=& sum _k=0^2 P(sqrtX=kcap Y=k)\
&=& sum _k=0^2 P(sqrtX=k)P(Y=k)\
&=& sum _k=0^2 P(X=k^2)P(Y=k)\
endeqnarray*
answered Jul 14 at 16:27
greedoid
26.6k93574
add a comment |Â
up vote
3
down vote
Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $sqrtX = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate
$$
P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)
$$
answered Jul 14 at 16:25
Brian Tung
25.3k32453
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
begineqnarray* P(sqrtX=Y) &=& sum _k=0^2 P(sqrtX=kcap Y=k)\
&=& sum _k=0^2 P(sqrtX=k)P(Y=k)\
&=& sum _k=0^2 P(X=k^2)P(Y=k)\
endeqnarray*
answered Jul 14 at 16:27
greedoid
26.6k93574
add a comment |Â
up vote
1
down vote
accepted
begineqnarray* P(sqrtX=Y) &=& sum _k=0^2 P(sqrtX=kcap Y=k)\
&=& sum _k=0^2 P(sqrtX=k)P(Y=k)\
&=& sum _k=0^2 P(X=k^2)P(Y=k)\
endeqnarray*
answered Jul 14 at 16:27
greedoid
26.6k93574
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
begineqnarray* P(sqrtX=Y) &=& sum _k=0^2 P(sqrtX=kcap Y=k)\
&=& sum _k=0^2 P(sqrtX=k)P(Y=k)\
&=& sum _k=0^2 P(X=k^2)P(Y=k)\
endeqnarray*
answered Jul 14 at 16:27
greedoid
26.6k93574
begineqnarray* P(sqrtX=Y) &=& sum _k=0^2 P(sqrtX=kcap Y=k)\
&=& sum _k=0^2 P(sqrtX=k)P(Y=k)\
&=& sum _k=0^2 P(X=k^2)P(Y=k)\
endeqnarray*
answered Jul 14 at 16:27
greedoid
26.6k93574
answered Jul 14 at 16:27
greedoid
26.6k93574
answered Jul 14 at 16:27
greedoid
26.6k93574
answered Jul 14 at 16:27
greedoid
26.6k93574
26.6k93574
add a comment |Â
add a comment |Â
up vote
3
down vote
Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $sqrtX = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate
$$
P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)
$$
answered Jul 14 at 16:25
Brian Tung
25.3k32453
add a comment |Â
up vote
3
down vote
Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $sqrtX = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate
$$
P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)
$$
answered Jul 14 at 16:25
Brian Tung
25.3k32453
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $sqrtX = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate
$$
P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)
$$
answered Jul 14 at 16:25
Brian Tung
25.3k32453
Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $sqrtX = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate
$$
P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)
$$
answered Jul 14 at 16:25
Brian Tung
25.3k32453
answered Jul 14 at 16:25
Brian Tung
25.3k32453
answered Jul 14 at 16:25
Brian Tung
25.3k32453
answered Jul 14 at 16:25
Brian Tung
25.3k32453
25.3k32453
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2851726%2fhow-do-i-myself-compute-the-following-probability%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions.
– callculus
Jul 14 at 16:10
I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes
– Roy Rizk
Jul 14 at 16:15
You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered.
– callculus
Jul 14 at 16:20
In R:
i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3))returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulationmean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2))returns 0.066067, correct to three places.– BruceET
Jul 14 at 16:38