Discrete Probability: Random Variable Independent or Dependent?
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Question a):
You flip a fair coin $7$ times; these coin flips are independent of each other.
Define the random variables:
$X$ = the number of heads in these $7$ coin flips, and
$Y$ = the number of tails in these $7$ coin flips.
Are the random variables independent or not? Show why or why not?
Attempt:
I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.
{HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}
If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$
The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.
The Probability of just 1 head and 1 tail occurring together is $0$
So, they are not equal and hence dependent. Is this correct way?
Question b):
Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.
Define the random variables:
$X$ = $0$ if z is even and $1$ is z odd
$Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}
Are the random variables independent or not? Show why or why not?
Attempt:
$Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.
$Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets
$Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$
$frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$
They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?
probability probability-theory discrete-mathematics random-variables
asked Dec 1 '18 at 23:44
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Question a):
You flip a fair coin $7$ times; these coin flips are independent of each other.
Define the random variables:
$X$ = the number of heads in these $7$ coin flips, and
$Y$ = the number of tails in these $7$ coin flips.
Are the random variables independent or not? Show why or why not?
Attempt:
I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.
{HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}
If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$
The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.
The Probability of just 1 head and 1 tail occurring together is $0$
So, they are not equal and hence dependent. Is this correct way?
Question b):
Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.
Define the random variables:
$X$ = $0$ if z is even and $1$ is z odd
$Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}
Are the random variables independent or not? Show why or why not?
Attempt:
$Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.
$Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets
$Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$
$frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$
They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?
probability probability-theory discrete-mathematics random-variables
asked Dec 1 '18 at 23:44
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Question a):
You flip a fair coin $7$ times; these coin flips are independent of each other.
Define the random variables:
$X$ = the number of heads in these $7$ coin flips, and
$Y$ = the number of tails in these $7$ coin flips.
Are the random variables independent or not? Show why or why not?
Attempt:
I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.
{HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}
If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$
The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.
The Probability of just 1 head and 1 tail occurring together is $0$
So, they are not equal and hence dependent. Is this correct way?
Question b):
Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.
Define the random variables:
$X$ = $0$ if z is even and $1$ is z odd
$Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}
Are the random variables independent or not? Show why or why not?
Attempt:
$Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.
$Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets
$Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$
$frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$
They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?
probability probability-theory discrete-mathematics random-variables
asked Dec 1 '18 at 23:44
TobyToby
1577
$endgroup$
Question a):
You flip a fair coin $7$ times; these coin flips are independent of each other.
Define the random variables:
$X$ = the number of heads in these $7$ coin flips, and
$Y$ = the number of tails in these $7$ coin flips.
Are the random variables independent or not? Show why or why not?
Attempt:
I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.
{HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}
If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$
The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.
The Probability of just 1 head and 1 tail occurring together is $0$
So, they are not equal and hence dependent. Is this correct way?
Question b):
Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.
Define the random variables:
$X$ = $0$ if z is even and $1$ is z odd
$Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}
Are the random variables independent or not? Show why or why not?
Attempt:
$Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.
$Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets
$Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$
$frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$
They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?
probability probability-theory discrete-mathematics random-variables
probability probability-theory discrete-mathematics random-variables
asked Dec 1 '18 at 23:44
TobyToby
1577
asked Dec 1 '18 at 23:44
TobyToby
1577
asked Dec 1 '18 at 23:44
TobyToby
1577
asked Dec 1 '18 at 23:44
TobyToby
1577
asked Dec 1 '18 at 23:44
TobyToby
1577
1577
add a comment |
add a comment |
1 Answer
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The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"
a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)
b) Do the same analysis for your other cases.
Yes... you can also check this through:
Is $P(X|Y) = P(X)$.
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
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$begingroup$
The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"
a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)
b) Do the same analysis for your other cases.
Yes... you can also check this through:
Is $P(X|Y) = P(X)$.
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"
a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)
b) Do the same analysis for your other cases.
Yes... you can also check this through:
Is $P(X|Y) = P(X)$.
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"
a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)
b) Do the same analysis for your other cases.
Yes... you can also check this through:
Is $P(X|Y) = P(X)$.
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"
a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)
b) Do the same analysis for your other cases.
Yes... you can also check this through:
Is $P(X|Y) = P(X)$.
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
answered Dec 2 '18 at 0:05
David G. StorkDavid G. Stork
10.7k31332
10.7k31332
add a comment |
add a comment |
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