expected value of Consecutive numbers [duplicate]
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Probability and expectancy problem [closed]
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We choose randomly a subset $mathit S$ of size 25 from the set ${ 1,2,...,100}$ , what is the expected value of number of Consecutive numbers in $mathit S$?
Consecutive numbers in $mathit S$: is a pair ${ i,i+1}$ where i$in$$mathit S$ and i+1$in$$mathit S$.
probability expected-value
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
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marked as duplicate by StubbornAtom, amWhy, A. Pongrácz, Did, Leucippus Dec 11 '18 at 0:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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This question already has an answer here:
Probability and expectancy problem [closed]
2 answers
We choose randomly a subset $mathit S$ of size 25 from the set ${ 1,2,...,100}$ , what is the expected value of number of Consecutive numbers in $mathit S$?
Consecutive numbers in $mathit S$: is a pair ${ i,i+1}$ where i$in$$mathit S$ and i+1$in$$mathit S$.
probability expected-value
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
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marked as duplicate by StubbornAtom, amWhy, A. Pongrácz, Did, Leucippus Dec 11 '18 at 0:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
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– JMoravitz
Dec 8 '18 at 16:52
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it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
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– MathStudent
Dec 8 '18 at 17:17
add a comment |
$begingroup$
This question already has an answer here:
Probability and expectancy problem [closed]
2 answers
We choose randomly a subset $mathit S$ of size 25 from the set ${ 1,2,...,100}$ , what is the expected value of number of Consecutive numbers in $mathit S$?
Consecutive numbers in $mathit S$: is a pair ${ i,i+1}$ where i$in$$mathit S$ and i+1$in$$mathit S$.
probability expected-value
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
$endgroup$
This question already has an answer here:
Probability and expectancy problem [closed]
2 answers
We choose randomly a subset $mathit S$ of size 25 from the set ${ 1,2,...,100}$ , what is the expected value of number of Consecutive numbers in $mathit S$?
Consecutive numbers in $mathit S$: is a pair ${ i,i+1}$ where i$in$$mathit S$ and i+1$in$$mathit S$.
This question already has an answer here:
Probability and expectancy problem [closed]
2 answers
probability expected-value
probability expected-value
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
edited Dec 8 '18 at 17:07
MathStudent
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
asked Dec 8 '18 at 16:42
MathStudentMathStudent
284
284
marked as duplicate by StubbornAtom, amWhy, A. Pongrácz, Did, Leucippus Dec 11 '18 at 0:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by StubbornAtom, amWhy, A. Pongrácz, Did, Leucippus Dec 11 '18 at 0:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
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– JMoravitz
Dec 8 '18 at 16:52
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it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
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– MathStudent
Dec 8 '18 at 17:17
add a comment |
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Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
$endgroup$
– JMoravitz
Dec 8 '18 at 16:52
$begingroup$
it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
$endgroup$
– MathStudent
Dec 8 '18 at 17:17
$begingroup$
Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
$endgroup$
– JMoravitz
Dec 8 '18 at 16:52
$begingroup$
Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
$endgroup$
– JMoravitz
Dec 8 '18 at 16:52
$begingroup$
it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
$endgroup$
– MathStudent
Dec 8 '18 at 17:17
$begingroup$
it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
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– MathStudent
Dec 8 '18 at 17:17
add a comment |
1 Answer
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Linearity of expectation is the key. There are 99 consecutive pairs, each one having a $frac{binom{98}{23}}{binom{100}{25}}=frac{25×24}{100×99}$ chance of being selected. Thus the expected number of consecutive pairs is $frac{25×24×99}{100×99}=6$.
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Linearity of expectation is the key. There are 99 consecutive pairs, each one having a $frac{binom{98}{23}}{binom{100}{25}}=frac{25×24}{100×99}$ chance of being selected. Thus the expected number of consecutive pairs is $frac{25×24×99}{100×99}=6$.
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
Linearity of expectation is the key. There are 99 consecutive pairs, each one having a $frac{binom{98}{23}}{binom{100}{25}}=frac{25×24}{100×99}$ chance of being selected. Thus the expected number of consecutive pairs is $frac{25×24×99}{100×99}=6$.
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
Linearity of expectation is the key. There are 99 consecutive pairs, each one having a $frac{binom{98}{23}}{binom{100}{25}}=frac{25×24}{100×99}$ chance of being selected. Thus the expected number of consecutive pairs is $frac{25×24×99}{100×99}=6$.
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
Linearity of expectation is the key. There are 99 consecutive pairs, each one having a $frac{binom{98}{23}}{binom{100}{25}}=frac{25×24}{100×99}$ chance of being selected. Thus the expected number of consecutive pairs is $frac{25×24×99}{100×99}=6$.
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:07
Parcly TaxelParcly Taxel
41.8k1372101
41.8k1372101
add a comment |
add a comment |
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$begingroup$
Hint: What is the probability that both $1$ and $2$ are in your randomly chosen subset?
$endgroup$
– JMoravitz
Dec 8 '18 at 16:52
$begingroup$
it is:$frac{binom{98}{23}}{binom{100}{25}}$ as Parcly mentioned below,that is the probability of any consecutive pairs not only 1 and 2.If we multiply that by number of possible consecutive pairs we will get the expected value we are looking for.@JMoravitz
$endgroup$
– MathStudent
Dec 8 '18 at 17:17