Probability of not drawing an element of a certain element in an “binomial drawing” experiment
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I randomly draw 3 elements from a population of 100. What is the probability of a certain element in my whole population NOT being including in my drawing?
I know I can draw $C^3_{100} = 161.700$ different ways but the probability mentioned above I cannot compute?
probability
asked Nov 22 at 18:14
k.dkhk
1628
add a comment |
up vote
0
down vote
favorite
I randomly draw 3 elements from a population of 100. What is the probability of a certain element in my whole population NOT being including in my drawing?
I know I can draw $C^3_{100} = 161.700$ different ways but the probability mentioned above I cannot compute?
probability
asked Nov 22 at 18:14
k.dkhk
1628
This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I randomly draw 3 elements from a population of 100. What is the probability of a certain element in my whole population NOT being including in my drawing?
I know I can draw $C^3_{100} = 161.700$ different ways but the probability mentioned above I cannot compute?
probability
asked Nov 22 at 18:14
k.dkhk
1628
I randomly draw 3 elements from a population of 100. What is the probability of a certain element in my whole population NOT being including in my drawing?
I know I can draw $C^3_{100} = 161.700$ different ways but the probability mentioned above I cannot compute?
probability
probability
asked Nov 22 at 18:14
k.dkhk
1628
asked Nov 22 at 18:14
k.dkhk
1628
asked Nov 22 at 18:14
k.dkhk
1628
asked Nov 22 at 18:14
k.dkhk
1628
asked Nov 22 at 18:14
k.dkhk
1628
1628
This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29
add a comment |
This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29
This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29
add a comment |
1 Answer
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P(a certain element is not in draw/sample)=$frac{{99choose3}}{{100choose3}}=frac{99times98times97}{100times99times98}=97/100.$
answered Nov 22 at 18:48
John_Wick
1,199111
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
P(a certain element is not in draw/sample)=$frac{{99choose3}}{{100choose3}}=frac{99times98times97}{100times99times98}=97/100.$
answered Nov 22 at 18:48
John_Wick
1,199111
add a comment |
up vote
0
down vote
accepted
P(a certain element is not in draw/sample)=$frac{{99choose3}}{{100choose3}}=frac{99times98times97}{100times99times98}=97/100.$
answered Nov 22 at 18:48
John_Wick
1,199111
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
P(a certain element is not in draw/sample)=$frac{{99choose3}}{{100choose3}}=frac{99times98times97}{100times99times98}=97/100.$
answered Nov 22 at 18:48
John_Wick
1,199111
P(a certain element is not in draw/sample)=$frac{{99choose3}}{{100choose3}}=frac{99times98times97}{100times99times98}=97/100.$
answered Nov 22 at 18:48
John_Wick
1,199111
answered Nov 22 at 18:48
John_Wick
1,199111
answered Nov 22 at 18:48
John_Wick
1,199111
answered Nov 22 at 18:48
John_Wick
1,199111
1,199111
add a comment |
add a comment |
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This looks a lot like a "homework" question but I am actually exploring a Scenario optimization model in which I need to compute these kinds of probabilites
– k.dkhk
Nov 22 at 18:16
If you cannot draw the 'certain element', then the favourable draws are $binom{99}{3}$.
– Daniel
Nov 22 at 18:27
This follows a simple case of a hypergeometric distribution where $K=1$ and $k=0$.
– JMoravitz
Nov 22 at 18:29