Probability and expectancy problem [closed]

-1

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?

edited Dec 5 at 18:46

greedoid

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asked Dec 5 at 18:35

user610402

214

closed as off-topic by amWhy, user21820, RRL, Lord_Farin, Did Dec 19 at 18:20

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Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56

add a comment |

-1

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?

edited Dec 5 at 18:46

greedoid

37.5k114794

asked Dec 5 at 18:35

user610402

214

closed as off-topic by amWhy, user21820, RRL, Lord_Farin, Did Dec 19 at 18:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user21820, RRL, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.

Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56

add a comment |

-1

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?

edited Dec 5 at 18:46

greedoid

37.5k114794

asked Dec 5 at 18:35

user610402

214

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

37.5k114794

asked Dec 5 at 18:35

user610402

214

edited Dec 5 at 18:46

greedoid

37.5k114794

asked Dec 5 at 18:35

user610402

214

edited Dec 5 at 18:46

greedoid

37.5k114794

edited Dec 5 at 18:46

greedoid

37.5k114794

edited Dec 5 at 18:46

greedoid

37.5k114794

asked Dec 5 at 18:35

user610402

214

asked Dec 5 at 18:35

user610402

214

asked Dec 5 at 18:35

user610402

214

closed as off-topic by amWhy, user21820, RRL, Lord_Farin, Did Dec 19 at 18:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user21820, RRL, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, user21820, RRL, Lord_Farin, Did Dec 19 at 18:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user21820, RRL, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.

Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56

add a comment |

Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56

Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56

add a comment |

2 Answers
2

active

oldest

votes

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

40.2k645107

add a comment |

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 $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

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

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

40.2k645107

add a comment |

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

40.2k645107

add a comment |

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

40.2k645107

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

40.2k645107

answered Dec 5 at 18:52

leonbloy

40.2k645107

answered Dec 5 at 18:52

leonbloy

40.2k645107

answered Dec 5 at 18:52

leonbloy

40.2k645107

add a comment |

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 $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

Is that better @copper.hat?
– greedoid
Dec 5 at 18:44

Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49

add a comment |

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 $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

Is that better @copper.hat?
– greedoid
Dec 5 at 18:44

Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49

add a comment |

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 $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

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 $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

edited Dec 5 at 18:59

answered Dec 5 at 18:39

greedoid

37.5k114794

answered Dec 5 at 18:39

greedoid

37.5k114794

answered Dec 5 at 18:39

greedoid

37.5k114794

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

Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49

add a comment |

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