The probability of choosing $j$ balls
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Suppose we have 9 balls that can be chosen with independent probabilities $p_1 = frac{1}{10}, p_2 = frac{2}{10}, p_3 = frac{3}{10}, p_4 = frac{1}{10}, p_5 = 0, p_6 = frac{1}{2}, p_7 = frac{1}{10}, p_8 = 0, p_9 = frac{2}{10}$, where $p_i$ is the probability with which the $i$th ball can be cosen.
I can't calculate the probability of choosing $j$ balls, where $0 le j le 9$ is arbitrary. Can someone help me, please?
probability probability-theory
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
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add a comment |
$begingroup$
Suppose we have 9 balls that can be chosen with independent probabilities $p_1 = frac{1}{10}, p_2 = frac{2}{10}, p_3 = frac{3}{10}, p_4 = frac{1}{10}, p_5 = 0, p_6 = frac{1}{2}, p_7 = frac{1}{10}, p_8 = 0, p_9 = frac{2}{10}$, where $p_i$ is the probability with which the $i$th ball can be cosen.
I can't calculate the probability of choosing $j$ balls, where $0 le j le 9$ is arbitrary. Can someone help me, please?
probability probability-theory
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
$endgroup$
$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49
add a comment |
$begingroup$
Suppose we have 9 balls that can be chosen with independent probabilities $p_1 = frac{1}{10}, p_2 = frac{2}{10}, p_3 = frac{3}{10}, p_4 = frac{1}{10}, p_5 = 0, p_6 = frac{1}{2}, p_7 = frac{1}{10}, p_8 = 0, p_9 = frac{2}{10}$, where $p_i$ is the probability with which the $i$th ball can be cosen.
I can't calculate the probability of choosing $j$ balls, where $0 le j le 9$ is arbitrary. Can someone help me, please?
probability probability-theory
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
$endgroup$
Suppose we have 9 balls that can be chosen with independent probabilities $p_1 = frac{1}{10}, p_2 = frac{2}{10}, p_3 = frac{3}{10}, p_4 = frac{1}{10}, p_5 = 0, p_6 = frac{1}{2}, p_7 = frac{1}{10}, p_8 = 0, p_9 = frac{2}{10}$, where $p_i$ is the probability with which the $i$th ball can be cosen.
I can't calculate the probability of choosing $j$ balls, where $0 le j le 9$ is arbitrary. Can someone help me, please?
probability probability-theory
probability probability-theory
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
asked Dec 19 '18 at 17:43
g.pomegranateg.pomegranate
1,066517
1,066517
$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49
add a comment |
$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49
$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49
add a comment |
1 Answer
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$begingroup$
Probability is defined for a specific outcome. Choosing $j$ balls is an experiment rather than an outcome. You can ask probability for an outcome like first ball being ball-1 and second ball being ball-7, etc. I don't understand what you exactly mean when you say probability of choosing $j$ balls.
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
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add a comment |
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1 Answer
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1 Answer
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active
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votes
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votes
$begingroup$
Probability is defined for a specific outcome. Choosing $j$ balls is an experiment rather than an outcome. You can ask probability for an outcome like first ball being ball-1 and second ball being ball-7, etc. I don't understand what you exactly mean when you say probability of choosing $j$ balls.
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
$endgroup$
add a comment |
$begingroup$
Probability is defined for a specific outcome. Choosing $j$ balls is an experiment rather than an outcome. You can ask probability for an outcome like first ball being ball-1 and second ball being ball-7, etc. I don't understand what you exactly mean when you say probability of choosing $j$ balls.
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
$endgroup$
add a comment |
$begingroup$
Probability is defined for a specific outcome. Choosing $j$ balls is an experiment rather than an outcome. You can ask probability for an outcome like first ball being ball-1 and second ball being ball-7, etc. I don't understand what you exactly mean when you say probability of choosing $j$ balls.
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
$endgroup$
Probability is defined for a specific outcome. Choosing $j$ balls is an experiment rather than an outcome. You can ask probability for an outcome like first ball being ball-1 and second ball being ball-7, etc. I don't understand what you exactly mean when you say probability of choosing $j$ balls.
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:45
Kshitij MishraKshitij Mishra
413
413
add a comment |
add a comment |
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$begingroup$
I don't see an easier way than to consider all the possible subsets of a given size. Of course there are some helpful symmetries, and for large $j$ you can work from the complement, but it's still messy. There are effectively seven balls, and $binom 74=35$ after all.
$endgroup$
– lulu
Dec 19 '18 at 17:47
$begingroup$
You can best exploit the symmetries by counting how many of the $frac 1{10}$ balls are chosen, then how many of the $frac 2{10}$ balls.
$endgroup$
– lulu
Dec 19 '18 at 17:49