Mixing
Probability of pulling out socks.
$begingroup$
A math task, where I've gotten mixed answers has occupied my mind, it goes like this.
A boy has his sister organize 7 pairs of socks into 2 different drawers.
Electricity goes out and the boy has to take out a pair of identical socks, what are the odds and why?
In my head, if you take a sock out from the first drawer, it doesn't affect the odds of taking out the identical pair from the 2nd one, so it should be 1/7, but people are saying it's 1/49, cause it matters which sock he takes from the first drawer, the question I have is, what is the correct answer and why?
probability
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
$endgroup$
add a comment |
$begingroup$
A math task, where I've gotten mixed answers has occupied my mind, it goes like this.
A boy has his sister organize 7 pairs of socks into 2 different drawers.
Electricity goes out and the boy has to take out a pair of identical socks, what are the odds and why?
In my head, if you take a sock out from the first drawer, it doesn't affect the odds of taking out the identical pair from the 2nd one, so it should be 1/7, but people are saying it's 1/49, cause it matters which sock he takes from the first drawer, the question I have is, what is the correct answer and why?
probability
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
$endgroup$
1
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
1
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40
add a comment |
$begingroup$
A math task, where I've gotten mixed answers has occupied my mind, it goes like this.
A boy has his sister organize 7 pairs of socks into 2 different drawers.
Electricity goes out and the boy has to take out a pair of identical socks, what are the odds and why?
In my head, if you take a sock out from the first drawer, it doesn't affect the odds of taking out the identical pair from the 2nd one, so it should be 1/7, but people are saying it's 1/49, cause it matters which sock he takes from the first drawer, the question I have is, what is the correct answer and why?
probability
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
$endgroup$
A math task, where I've gotten mixed answers has occupied my mind, it goes like this.
A boy has his sister organize 7 pairs of socks into 2 different drawers.
Electricity goes out and the boy has to take out a pair of identical socks, what are the odds and why?
In my head, if you take a sock out from the first drawer, it doesn't affect the odds of taking out the identical pair from the 2nd one, so it should be 1/7, but people are saying it's 1/49, cause it matters which sock he takes from the first drawer, the question I have is, what is the correct answer and why?
probability
probability
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
asked Jan 29 at 8:15
Darius UscinsDarius Uscins
61
61
1
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
1
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40
add a comment |
1
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
1
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40
1
1
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
1
1
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You mentioned taking out one pair, not a specific one. So it does not matter which one you take from the 1st drawer. You still have a chance of 1:7 to get a matching one.
If let's say the pairs would have different colors and you'd need to get the red pair, in that case there would be an initial 1:7 to get a red from the 1st drawer combined with the same probability to get the red from the 2nd drawer, so in total it would make the probability 1:49 go get the specific pair.
answered Jan 29 at 9:11
OvermindOvermind
1133
$endgroup$
add a comment |
$begingroup$
From what I've gathered.
If you have sock pairs split up and put in to two different drawers, and they are distinguishable from one another then you have two answers.
- If a defined pair is required to be taken out it is 1/49th,
- If you just need to get a pair out it is 1/7th
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
oldest
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$begingroup$
You mentioned taking out one pair, not a specific one. So it does not matter which one you take from the 1st drawer. You still have a chance of 1:7 to get a matching one.
If let's say the pairs would have different colors and you'd need to get the red pair, in that case there would be an initial 1:7 to get a red from the 1st drawer combined with the same probability to get the red from the 2nd drawer, so in total it would make the probability 1:49 go get the specific pair.
answered Jan 29 at 9:11
OvermindOvermind
1133
$endgroup$
add a comment |
$begingroup$
You mentioned taking out one pair, not a specific one. So it does not matter which one you take from the 1st drawer. You still have a chance of 1:7 to get a matching one.
If let's say the pairs would have different colors and you'd need to get the red pair, in that case there would be an initial 1:7 to get a red from the 1st drawer combined with the same probability to get the red from the 2nd drawer, so in total it would make the probability 1:49 go get the specific pair.
answered Jan 29 at 9:11
OvermindOvermind
1133
$endgroup$
add a comment |
$begingroup$
You mentioned taking out one pair, not a specific one. So it does not matter which one you take from the 1st drawer. You still have a chance of 1:7 to get a matching one.
If let's say the pairs would have different colors and you'd need to get the red pair, in that case there would be an initial 1:7 to get a red from the 1st drawer combined with the same probability to get the red from the 2nd drawer, so in total it would make the probability 1:49 go get the specific pair.
answered Jan 29 at 9:11
OvermindOvermind
1133
$endgroup$
You mentioned taking out one pair, not a specific one. So it does not matter which one you take from the 1st drawer. You still have a chance of 1:7 to get a matching one.
If let's say the pairs would have different colors and you'd need to get the red pair, in that case there would be an initial 1:7 to get a red from the 1st drawer combined with the same probability to get the red from the 2nd drawer, so in total it would make the probability 1:49 go get the specific pair.
answered Jan 29 at 9:11
OvermindOvermind
1133
answered Jan 29 at 9:11
OvermindOvermind
1133
answered Jan 29 at 9:11
OvermindOvermind
1133
answered Jan 29 at 9:11
OvermindOvermind
1133
1133
add a comment |
add a comment |
$begingroup$
From what I've gathered.
If you have sock pairs split up and put in to two different drawers, and they are distinguishable from one another then you have two answers.
- If a defined pair is required to be taken out it is 1/49th,
- If you just need to get a pair out it is 1/7th
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
$endgroup$
add a comment |
$begingroup$
From what I've gathered.
If you have sock pairs split up and put in to two different drawers, and they are distinguishable from one another then you have two answers.
- If a defined pair is required to be taken out it is 1/49th,
- If you just need to get a pair out it is 1/7th
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
$endgroup$
add a comment |
$begingroup$
From what I've gathered.
If you have sock pairs split up and put in to two different drawers, and they are distinguishable from one another then you have two answers.
- If a defined pair is required to be taken out it is 1/49th,
- If you just need to get a pair out it is 1/7th
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
$endgroup$
From what I've gathered.
If you have sock pairs split up and put in to two different drawers, and they are distinguishable from one another then you have two answers.
- If a defined pair is required to be taken out it is 1/49th,
- If you just need to get a pair out it is 1/7th
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
answered Jan 29 at 8:55
Darius UscinsDarius Uscins
61
61
add a comment |
add a comment |
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1
$begingroup$
If the question refers to the probability of getting an undefined pair of identical socks, the answer is indeed $frac{1}{7}.$ Only if the question refers to the probability of getting a specific pair of socks, the answer is $frac{1}{49}.$
$endgroup$
– jvdhooft
Jan 29 at 8:32
1
$begingroup$
The seven pairs are distinguishable from each other? and from each pair, one sock goes into one drawer, and the other sock goes into the other drawer?
$endgroup$
– Gerry Myerson
Jan 29 at 8:35
$begingroup$
@jvdhooft So if the pair of socks is defined, it's 1 49th, but if it's not defined it is 1 7th okay. At GerryMyerson they are distinguishable from each other and pairs are separated in to two different drawers.
$endgroup$
– Darius Uscins
Jan 29 at 8:38
$begingroup$
@DariusUscins Assuming that socks are distinguishable (e.g., by color), and that socks belonging to the same pair are put in different drawers, the above is indeed correct.
$endgroup$
– jvdhooft
Jan 29 at 8:40