The salesman randomly pulled out $8$ socks out of $14$ pairs of socks. The 2019 Stack Overflow Developer Survey Results Are InProbability : A bag contains 12 pair socks . Four socks are picked up at random. Find the probability that there is at least one pair.probability for socks pairsArrangements of n pairs of socks on a clotheslineThis is a variation of the colored socks in a drawer problem.From a bag containing $10$ pairs of socks, how many must a person pull out to ensure that they get at least $2$ matching pairs of socks?What is the smallest number of socks you should pull out so that you can be assured that you will have at least one pair of matching socks?The probability to match pair of socksSampling four socks from $n$ distinct pairs of socksWork out the probability that my odd socks are specifically targeted by sock stealing gnomes.Probability of pulling out socks.
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The salesman randomly pulled out $8$ socks out of $14$ pairs of socks.
The 2019 Stack Overflow Developer Survey Results Are InProbability : A bag contains 12 pair socks . Four socks are picked up at random. Find the probability that there is at least one pair.probability for socks pairsArrangements of n pairs of socks on a clotheslineThis is a variation of the colored socks in a drawer problem.From a bag containing $10$ pairs of socks, how many must a person pull out to ensure that they get at least $2$ matching pairs of socks?What is the smallest number of socks you should pull out so that you can be assured that you will have at least one pair of matching socks?The probability to match pair of socksSampling four socks from $n$ distinct pairs of socksWork out the probability that my odd socks are specifically targeted by sock stealing gnomes.Probability of pulling out socks.
$begingroup$
The salesman randomly pulled out $8$ socks out of $14$ pairs of socks. Calculate the probability of a case where there is at least one pair of socks among the pulled ones.
My question is: should I be calculating as if all socks were different or as if they form multiset of $14$ elements?
probability combinatorics
asked Mar 23 at 17:30
user560461user560461
1568
$endgroup$
add a comment |
$begingroup$
The salesman randomly pulled out $8$ socks out of $14$ pairs of socks. Calculate the probability of a case where there is at least one pair of socks among the pulled ones.
My question is: should I be calculating as if all socks were different or as if they form multiset of $14$ elements?
probability combinatorics
asked Mar 23 at 17:30
user560461user560461
1568
$endgroup$
$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48
add a comment |
$begingroup$
The salesman randomly pulled out $8$ socks out of $14$ pairs of socks. Calculate the probability of a case where there is at least one pair of socks among the pulled ones.
My question is: should I be calculating as if all socks were different or as if they form multiset of $14$ elements?
probability combinatorics
asked Mar 23 at 17:30
user560461user560461
1568
$endgroup$
The salesman randomly pulled out $8$ socks out of $14$ pairs of socks. Calculate the probability of a case where there is at least one pair of socks among the pulled ones.
My question is: should I be calculating as if all socks were different or as if they form multiset of $14$ elements?
probability combinatorics
probability combinatorics
asked Mar 23 at 17:30
user560461user560461
1568
asked Mar 23 at 17:30
user560461user560461
1568
asked Mar 23 at 17:30
user560461user560461
1568
asked Mar 23 at 17:30
user560461user560461
1568
asked Mar 23 at 17:30
user560461user560461
1568
1568
$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48
add a comment |
$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48
$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are $binom288$ choices of sock, of which $2^8binom148$ don't get both the socks from any one pair, so you want $$1-frac2^8binom148binom288=1-frac2^8cdot14cdot13cdot12cdot11cdot10cdot9cdot8cdot728cdot27cdot26cdot25cdot24cdot23cdot22cdot21=1-frac2^4cdot2cdot83cdot5cdot23cdot3=1-frac2561035=frac7791035.$$
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are $binom288$ choices of sock, of which $2^8binom148$ don't get both the socks from any one pair, so you want $$1-frac2^8binom148binom288=1-frac2^8cdot14cdot13cdot12cdot11cdot10cdot9cdot8cdot728cdot27cdot26cdot25cdot24cdot23cdot22cdot21=1-frac2^4cdot2cdot83cdot5cdot23cdot3=1-frac2561035=frac7791035.$$
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
There are $binom288$ choices of sock, of which $2^8binom148$ don't get both the socks from any one pair, so you want $$1-frac2^8binom148binom288=1-frac2^8cdot14cdot13cdot12cdot11cdot10cdot9cdot8cdot728cdot27cdot26cdot25cdot24cdot23cdot22cdot21=1-frac2^4cdot2cdot83cdot5cdot23cdot3=1-frac2561035=frac7791035.$$
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
There are $binom288$ choices of sock, of which $2^8binom148$ don't get both the socks from any one pair, so you want $$1-frac2^8binom148binom288=1-frac2^8cdot14cdot13cdot12cdot11cdot10cdot9cdot8cdot728cdot27cdot26cdot25cdot24cdot23cdot22cdot21=1-frac2^4cdot2cdot83cdot5cdot23cdot3=1-frac2561035=frac7791035.$$
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
$endgroup$
There are $binom288$ choices of sock, of which $2^8binom148$ don't get both the socks from any one pair, so you want $$1-frac2^8binom148binom288=1-frac2^8cdot14cdot13cdot12cdot11cdot10cdot9cdot8cdot728cdot27cdot26cdot25cdot24cdot23cdot22cdot21=1-frac2^4cdot2cdot83cdot5cdot23cdot3=1-frac2561035=frac7791035.$$
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
answered Mar 23 at 18:52
J.G.J.G.
33.2k23252
33.2k23252
add a comment |
add a comment |
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$begingroup$
Work from the complement. The first choice is free, the second can be any of $26$. Third can be any of $24$, and so on.
$endgroup$
– lulu
Mar 23 at 17:37
$begingroup$
@lulu that means that all paired socks are different? I kinda feel like you can not separate which is left and which is right, that is why I asked this question in first place.
$endgroup$
– user560461
Mar 23 at 17:45
$begingroup$
Not sure what you mean. I'm just saying you have to avoid the pairs that were previously chosen.
$endgroup$
– lulu
Mar 23 at 17:47
$begingroup$
Let's say he selects two choices. Then the probability that they come from different pairs is just $frac 2627$, since there are $27$ socks left to choose from and only one bad choice. That's all I'm saying.
$endgroup$
– lulu
Mar 23 at 17:48