Mixing
understanding the reason a calculation occured in a joint dependent experiment
$begingroup$
i don't understand the multipication by 1/2 in the calculation. i'll provide description of the problem and what i did, in the calculation i do not understand.
there are 2 cages; in cage 1 there are one male and one female of some species. in cage 2 there are two females and two females of the same species. randomly, we choose a cage and take two species out of it.we define N - the number of cage the species were taken from. Z - number of male species taken.
i need to find the joint probability function of N and Z and to check if they are dependent or not.
the part i am having problem with:
so we can look at it like a bernoli trial, if cage one was chosen the only option is there is only one male(Z is hypergeometric as i understand) so, if we choose N=2 and Z=0, then the probability $P{n=2,Z=0}$ is $P{n=2,Z=0}=P{n=2|Z=0}P{n=2}=frac{binom20binom22}{binom42} * 1/2$. but i don't understand the 1/2: is it because i chose 1 cage out of 2 or is it something different? sorry for the "lame" question, i am in the beginning of my learning curve.
thank you very much! (p.s: in my opinion N and Z are dependent, since P{N=i}P{Z=j} are not going to be equal to P{N=i,Z=j}, in my opinion).
probability probability-distributions
asked Jan 28 at 14:11
q123q123
75
$endgroup$
add a comment |
$begingroup$
i don't understand the multipication by 1/2 in the calculation. i'll provide description of the problem and what i did, in the calculation i do not understand.
there are 2 cages; in cage 1 there are one male and one female of some species. in cage 2 there are two females and two females of the same species. randomly, we choose a cage and take two species out of it.we define N - the number of cage the species were taken from. Z - number of male species taken.
i need to find the joint probability function of N and Z and to check if they are dependent or not.
the part i am having problem with:
so we can look at it like a bernoli trial, if cage one was chosen the only option is there is only one male(Z is hypergeometric as i understand) so, if we choose N=2 and Z=0, then the probability $P{n=2,Z=0}$ is $P{n=2,Z=0}=P{n=2|Z=0}P{n=2}=frac{binom20binom22}{binom42} * 1/2$. but i don't understand the 1/2: is it because i chose 1 cage out of 2 or is it something different? sorry for the "lame" question, i am in the beginning of my learning curve.
thank you very much! (p.s: in my opinion N and Z are dependent, since P{N=i}P{Z=j} are not going to be equal to P{N=i,Z=j}, in my opinion).
probability probability-distributions
asked Jan 28 at 14:11
q123q123
75
$endgroup$
add a comment |
$begingroup$
i don't understand the multipication by 1/2 in the calculation. i'll provide description of the problem and what i did, in the calculation i do not understand.
there are 2 cages; in cage 1 there are one male and one female of some species. in cage 2 there are two females and two females of the same species. randomly, we choose a cage and take two species out of it.we define N - the number of cage the species were taken from. Z - number of male species taken.
i need to find the joint probability function of N and Z and to check if they are dependent or not.
the part i am having problem with:
so we can look at it like a bernoli trial, if cage one was chosen the only option is there is only one male(Z is hypergeometric as i understand) so, if we choose N=2 and Z=0, then the probability $P{n=2,Z=0}$ is $P{n=2,Z=0}=P{n=2|Z=0}P{n=2}=frac{binom20binom22}{binom42} * 1/2$. but i don't understand the 1/2: is it because i chose 1 cage out of 2 or is it something different? sorry for the "lame" question, i am in the beginning of my learning curve.
thank you very much! (p.s: in my opinion N and Z are dependent, since P{N=i}P{Z=j} are not going to be equal to P{N=i,Z=j}, in my opinion).
probability probability-distributions
asked Jan 28 at 14:11
q123q123
75
$endgroup$
i don't understand the multipication by 1/2 in the calculation. i'll provide description of the problem and what i did, in the calculation i do not understand.
there are 2 cages; in cage 1 there are one male and one female of some species. in cage 2 there are two females and two females of the same species. randomly, we choose a cage and take two species out of it.we define N - the number of cage the species were taken from. Z - number of male species taken.
i need to find the joint probability function of N and Z and to check if they are dependent or not.
the part i am having problem with:
so we can look at it like a bernoli trial, if cage one was chosen the only option is there is only one male(Z is hypergeometric as i understand) so, if we choose N=2 and Z=0, then the probability $P{n=2,Z=0}$ is $P{n=2,Z=0}=P{n=2|Z=0}P{n=2}=frac{binom20binom22}{binom42} * 1/2$. but i don't understand the 1/2: is it because i chose 1 cage out of 2 or is it something different? sorry for the "lame" question, i am in the beginning of my learning curve.
thank you very much! (p.s: in my opinion N and Z are dependent, since P{N=i}P{Z=j} are not going to be equal to P{N=i,Z=j}, in my opinion).
probability probability-distributions
probability probability-distributions
asked Jan 28 at 14:11
q123q123
75
asked Jan 28 at 14:11
q123q123
75
asked Jan 28 at 14:11
q123q123
75
asked Jan 28 at 14:11
q123q123
75
asked Jan 28 at 14:11
q123q123
75
75
add a comment |
add a comment |
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