Probability that $1$ and $2in A$ where $Asubset X_n$
$begingroup$
Let $x_n={1,2,3,...,n}$ and let a subset $A$ of $X_n$ be chosen so that every pair of elements of $A$ differ by at least $3$. When $n=10$, let probability that $1in A=p$ and probability that $2in A=q$.
Find $p$ and $q$.
I am not being able to find either the sample space or decide how to find the number of events in $p$ and $q$.
How should I do it? Thanks in advance!
probability discrete-mathematics
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
$endgroup$
add a comment |
$begingroup$
Let $x_n={1,2,3,...,n}$ and let a subset $A$ of $X_n$ be chosen so that every pair of elements of $A$ differ by at least $3$. When $n=10$, let probability that $1in A=p$ and probability that $2in A=q$.
Find $p$ and $q$.
I am not being able to find either the sample space or decide how to find the number of events in $p$ and $q$.
How should I do it? Thanks in advance!
probability discrete-mathematics
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
$endgroup$
$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17
add a comment |
$begingroup$
Let $x_n={1,2,3,...,n}$ and let a subset $A$ of $X_n$ be chosen so that every pair of elements of $A$ differ by at least $3$. When $n=10$, let probability that $1in A=p$ and probability that $2in A=q$.
Find $p$ and $q$.
I am not being able to find either the sample space or decide how to find the number of events in $p$ and $q$.
How should I do it? Thanks in advance!
probability discrete-mathematics
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
$endgroup$
Let $x_n={1,2,3,...,n}$ and let a subset $A$ of $X_n$ be chosen so that every pair of elements of $A$ differ by at least $3$. When $n=10$, let probability that $1in A=p$ and probability that $2in A=q$.
Find $p$ and $q$.
I am not being able to find either the sample space or decide how to find the number of events in $p$ and $q$.
How should I do it? Thanks in advance!
probability discrete-mathematics
probability discrete-mathematics
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
edited Dec 6 '18 at 18:44
greedoid
46.4k1160118
46.4k1160118
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
asked Nov 1 '18 at 18:11
tatantatan
5,79762759
5,79762759
$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17
add a comment |
$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17
$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17
$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Lulu's hint is good and his formula is correct, but argument is not. Say subset is good if every two differ for at least $3$. So, if $s_n$ is a number of good subsets in ${1,2,...,n}$ then we have $$s_n = s_{n-3}+s_{n-1}$$
(that is, if $n$ is in a good subset then we must take the rest of elements in ${1,2,...,n-3}$ else we take elements from a set ${1,2,...,n-1}$).
Now let $a_n$ be a number of good subsets with $1$ in it. Then $a_n=s_{n-3}$
Since first $11$ terms is (starting with $s_0$) $$1,2,3,4,6,9,10,15,19,25,34$$ we have
$$p = {a_{10}over s_{10}}= {s_{7}over s_{10}} = {15over 34}$$
You can procede similary for $q$ (note that $b_n=a_{n-4}$)
edited Nov 2 '18 at 4:06
tatan
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
$endgroup$
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Lulu's hint is good and his formula is correct, but argument is not. Say subset is good if every two differ for at least $3$. So, if $s_n$ is a number of good subsets in ${1,2,...,n}$ then we have $$s_n = s_{n-3}+s_{n-1}$$
(that is, if $n$ is in a good subset then we must take the rest of elements in ${1,2,...,n-3}$ else we take elements from a set ${1,2,...,n-1}$).
Now let $a_n$ be a number of good subsets with $1$ in it. Then $a_n=s_{n-3}$
Since first $11$ terms is (starting with $s_0$) $$1,2,3,4,6,9,10,15,19,25,34$$ we have
$$p = {a_{10}over s_{10}}= {s_{7}over s_{10}} = {15over 34}$$
You can procede similary for $q$ (note that $b_n=a_{n-4}$)
edited Nov 2 '18 at 4:06
tatan
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
$endgroup$
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
add a comment |
$begingroup$
Lulu's hint is good and his formula is correct, but argument is not. Say subset is good if every two differ for at least $3$. So, if $s_n$ is a number of good subsets in ${1,2,...,n}$ then we have $$s_n = s_{n-3}+s_{n-1}$$
(that is, if $n$ is in a good subset then we must take the rest of elements in ${1,2,...,n-3}$ else we take elements from a set ${1,2,...,n-1}$).
Now let $a_n$ be a number of good subsets with $1$ in it. Then $a_n=s_{n-3}$
Since first $11$ terms is (starting with $s_0$) $$1,2,3,4,6,9,10,15,19,25,34$$ we have
$$p = {a_{10}over s_{10}}= {s_{7}over s_{10}} = {15over 34}$$
You can procede similary for $q$ (note that $b_n=a_{n-4}$)
edited Nov 2 '18 at 4:06
tatan
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
$endgroup$
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
add a comment |
$begingroup$
Lulu's hint is good and his formula is correct, but argument is not. Say subset is good if every two differ for at least $3$. So, if $s_n$ is a number of good subsets in ${1,2,...,n}$ then we have $$s_n = s_{n-3}+s_{n-1}$$
(that is, if $n$ is in a good subset then we must take the rest of elements in ${1,2,...,n-3}$ else we take elements from a set ${1,2,...,n-1}$).
Now let $a_n$ be a number of good subsets with $1$ in it. Then $a_n=s_{n-3}$
Since first $11$ terms is (starting with $s_0$) $$1,2,3,4,6,9,10,15,19,25,34$$ we have
$$p = {a_{10}over s_{10}}= {s_{7}over s_{10}} = {15over 34}$$
You can procede similary for $q$ (note that $b_n=a_{n-4}$)
edited Nov 2 '18 at 4:06
tatan
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
$endgroup$
Lulu's hint is good and his formula is correct, but argument is not. Say subset is good if every two differ for at least $3$. So, if $s_n$ is a number of good subsets in ${1,2,...,n}$ then we have $$s_n = s_{n-3}+s_{n-1}$$
(that is, if $n$ is in a good subset then we must take the rest of elements in ${1,2,...,n-3}$ else we take elements from a set ${1,2,...,n-1}$).
Now let $a_n$ be a number of good subsets with $1$ in it. Then $a_n=s_{n-3}$
Since first $11$ terms is (starting with $s_0$) $$1,2,3,4,6,9,10,15,19,25,34$$ we have
$$p = {a_{10}over s_{10}}= {s_{7}over s_{10}} = {15over 34}$$
You can procede similary for $q$ (note that $b_n=a_{n-4}$)
edited Nov 2 '18 at 4:06
tatan
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
edited Nov 2 '18 at 4:06
tatan
5,79762759
edited Nov 2 '18 at 4:06
tatan
5,79762759
edited Nov 2 '18 at 4:06
tatan
5,79762759
5,79762759
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
answered Nov 1 '18 at 20:05
greedoidgreedoid
46.4k1160118
46.4k1160118
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
add a comment |
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
$begingroup$
Is this not good enough answer?
$endgroup$
– greedoid
Dec 6 '18 at 18:46
add a comment |
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$begingroup$
Hint: work recursively. Let $s_n$ be the number of such subsets. Either $1$ is in the subset or not, so $s_n=s_{n-3}+s_{n-1}$.
$endgroup$
– lulu
Nov 1 '18 at 18:17