Probability and expectancy problem
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if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.4k114592
asked Dec 5 at 18:35
user610402
354
add a comment |
up vote
5
down vote
favorite
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.4k114592
asked Dec 5 at 18:35
user610402
354
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.4k114592
asked Dec 5 at 18:35
user610402
354
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
probability expected-value
edited Dec 5 at 18:46
greedoid
36.4k114592
asked Dec 5 at 18:35
user610402
354
edited Dec 5 at 18:46
greedoid
36.4k114592
asked Dec 5 at 18:35
user610402
354
edited Dec 5 at 18:46
greedoid
36.4k114592
edited Dec 5 at 18:46
greedoid
36.4k114592
edited Dec 5 at 18:46
greedoid
36.4k114592
36.4k114592
asked Dec 5 at 18:35
user610402
354
asked Dec 5 at 18:35
user610402
354
asked Dec 5 at 18:35
user610402
354
354
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday
add a comment |
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
39.9k645107
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.4k114592
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
39.9k645107
add a comment |
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
39.9k645107
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
39.9k645107
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
39.9k645107
answered Dec 5 at 18:52
leonbloy
39.9k645107
answered Dec 5 at 18:52
leonbloy
39.9k645107
answered Dec 5 at 18:52
leonbloy
39.9k645107
39.9k645107
add a comment |
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.4k114592
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.4k114592
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
up vote
7
down vote
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.4k114592
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.4k114592
edited Dec 5 at 18:59
answered Dec 5 at 18:39
greedoid
36.4k114592
answered Dec 5 at 18:39
greedoid
36.4k114592
answered Dec 5 at 18:39
greedoid
36.4k114592
36.4k114592
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
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Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
yesterday