The textbook answer was actually the same answer that I came up with during the video - just a little paraphrased. When taking out all socks one after another and laying them out in a line, you can get all possible permutations of the socks with equal probability. You can imagine all these permutations as lines one below another in a matrix. When we now permute the 1st with the 3rd column and the 2nd with the 4th column there are still all possible permutations represented. However, now the 3rd and 4th sock (tuesday's socks) are chosen before the 1st and the 2nd sock (monday's socks). Thus, you can arbitrarily relabel the days.
At first, I thought the answer wouldn't be 1/9, because I thought the different scenarios would result in a different final answer. However, this math problem was a bit of an illusion for me. Here is how I thought through it. Scenario #1: He selects a correct pair on day 1, so all remaining socks have a potential match for day 2. There's a 1/9 chance that he chose the correct match on day 1. That means there are 8 socks remaining. The first selection on the second day won't matter, so there are seven socks left. There is a 1 in 7 chance that he makes a correct match. Between days one and two, there is a 1 in 63 chance that he gets them both correct. Scenario #2: He makes an incorrect match on day 1. On day one, there is an 8 in 9 probability that he made an incorrect match. This means of the 8 socks remaining, only 6 of them have a match remaining in the drawer. This time, the first sock he chooses does matter. It must be one of the socks with a matching pair remaining, so there is a 6 out of 8 probability. That leaves 7 socks remaining, and the one he chooses must match. There is a 1 out of 7 probability. That makes a 6 out of 56 probability (6/8*1/7) that the second day results in a correct match. Altogether, there is an 8 out of 9 probability that the first day was an incorrect match and a 6 out of 56 probability (or 3 out of 28) that second day was correct. Thus makes an overall probability of 2 out 21 (or 6 out of 63) that the first day was incorrect but the second day was correct. Final answer: He could have a correct match on day two by being correct on day one (1 out of 63), and he could have a correct match on day two by being wrong on day 1 (6 out of 63). Together that makes 7 out of 63... otherwise known as 1 out of 9. The answer of 1/9 is correct, and using formulas is much more efficient.
On Monday he picks sock A and B. On Tuesday he picks the other sock from pair B. He now has 0 chance of getting a pair. The point is, your probability on days 2 onwards depend on what socks you picked on preceding days. Also is all socks were the same, then he has a 100% probability.
This is really interesting to see. Ever since i've learnt nCr notation I've basically forgotten how regular probability questions work lol. I got the same answer by just doing 5 x 1/10C2 which is 1/9.
if a matching pair is taken day 1, then 4 pairs remain, so Prob of picking another pair is 1/7 ( Pr Pick a sock = 1 , then Pr picking matching sock is 1/7 ) if non -matching pair is taken day 1, 3 matching pairs remain and 2 odd socks. Pr of picking either of those 2 odds socks Pr=2/8 - in which case it's impossible to find a pair Pr of picking a sock that still has a pair in the drawer Pr=6/8 having picked a sock with a potential pair, picking that pair Pr = 1/7 Pr of picking a pair on day2 = Pr picking a pair on day 1 ( 1/9) * 1/7 + not picking a pair day 1 , (8/9) * picking a sock that still has a pair in drawing (6/8) * picking that pair (1/7) =1/63 + 6/63 = 7/63 = 1/9 !
There are only 45 possible pairs, because pair AA and AA is the same (doesn't matter which A you picked first) - so possible pairs are 10*9 / 2. After that you get to 1/9 which is correct.
If Monday has 1/9 of getting a pair, it's 8/9 it was not. If there is no replacement then Tuesday has 8 left with a 8/9 chance that two of them don't have a pair. The first sock's favorable set then becomes from 8 to six. So it is 6/8. Or 3/4. Then the second sock has a chance of 1/7 chance to match since there are 7 socks left. 3/4 x 1/7 = 0.10714285714 or 3/28. IDK but it seems like 3/28 would be it. Maybe I am overthinking this.
Omg I literally just did this exact thing before reading your comment. I do believe you’re right. Yes you would get 3/28 as the probability of getting a pair on Tuesday. But then, you’d have to multiply that by the probably of not getting a pair on Monday which was 8/9. So 3/28 x 8/9 which equals 2/21.
@@caydenwaltersmusicYou then add that to the odds that the first day and second day were a match (1 out of 63) and you end up at 7/63 which is 1/9. It seems a bit counterintuitive at first, but the math checks out. The probability for day 2 matches the probability for day 1 irregardless of whether day 1 was a correct match or not.
There are 5 distinct pairs of socks, each pair of a different color. Assuming all arrangements are equally probable. Suppose you pick two socks at random. There are in all C(10,2) = 45, combinations but there are only 5 which are favourable to the desired event which means the probability is .11111..... .
Basically an answer that makes sense in theory but not in practice. I.E: Calculate the probability but don't assume that he's wearing the socks, he's just picking them. I feel like it's misleading and would punish students who think rationnaly such as that kid who said "what about washing?"
Ideally there are 36 possible combinations in which 6 are perfectly matched. Hence the probability of getting a matched pair is 6/36 otherwise 1/6. So why not 1/6??
2nd day, no 1st day match, 2/8 socks have you “drawing dead” after 1st draw. That feeling when you’re being taught how to think in terms of pokerr without realizing they’re being taught how to think in terms of poker….
Not sure if you are agreeing with the final outcome or not, but when you consider both scenarios for the first day (match vs no match), you end up with a 1 out of 9 probability for a 2nd day match.
I am afraid not. This is the probability of that desired outcome in exactly one trial of the experiment, not in seven trials of it. Week has nothing to do with it. In most finite probability problems you clearly identify the sample space of the possible outcomes in a single trial of the experiment and also those outcomes favourable to the desired event as a subset of the sample space and finally just take the ratio of their cardinalities. This is for simple events, for compound and complex events you will have to use other set theory as well as a bit more advanced probability theory concepts and principles. There must be a practical use of this theory for the advancement of human knowledge of the world we humans live in cause these can lead to a better understanding of the way this world truly scientifically works, with total perfection and precision, and also lead to a better appreciation of the underlying operational laws and principles governing the working of this world. It is certainly useful to us if we going to be in it. Or maybe we could simply sing, "We are the world". That may or may not cut it, depending on which tree you are talking about. Rather than simply say the quantum model is a non deterministic one for instance the seemingly random behavior of elementary particles could be seen as a probabilistic one, or in genes and vaccines the behavioral anamolies or abberations treated as probabilities. These principles can most certainly be grasped and applied in real life to solve real world problems. Although other than for my own personal use, with tremendous success, I have never really applied these principles in real world scenarios.
An honest question: If he only had enough socks for 5 days and not 7 sets of 5 days, would that change the probability in this case? I'm afraid he only has enough socks for one experiment. 😊 How do you see it? As someone with basic mathematical abilities, I am impressed with the way you think. I seem to remember reading an article once about the effect that reiterations have on the probability of an outcome. Thanks for renewing my interest.
This will probably not be answered, but my question is: If the question is without replacement, why is it that 1/9 applies on tuesday, wouldn't it be 1/7?
Because the probability of picking an acceptable 3rd sock is not necessarily 8/8 (1). Consider a case where on Monday you picked not matching pair. That means that on Tuesday the first sock you pick might not have a pair to pick anymore (as you picked it on Monday already). You need to consider both the case of picking a matching pair on Monday and picking non-matching pair on Monday and add the probabilities together. Probability to pick a match on both days is 1/63. Probability to pick a match only on Tuesday is 6/63. If you add those together you get 7/63 = 1/9.
Boeing lost everything in macas and designs boring engineers cause you have only an ba maths licence of twarts converts ts if di falled hi teachers of college algebras .
Your teachings is really great but did you build something new out of nowhere of course not because the complexes maths died not answer to the linearization of linears maths furthermore of complexes conundrums never stays the same as complexes systems cannot file the true nature of your self taught dusfuveried as a master can you explain why hiring list everything in your perfect linearisstion ion if slicing algebras algebras like a bra twarts if Boeing .
No offense but maybe don't call it "very easy", call his method difficult, and then say the "simple answer" is 5c2/10c2 when 1) most courses learn probability before nCr notation and 2) you're not even correct it's just 5/10c2.
Watching math for fun now !
The textbook answer was actually the same answer that I came up with during the video - just a little paraphrased.
When taking out all socks one after another and laying them out in a line, you can get all possible permutations of the socks with equal probability. You can imagine all these permutations as lines one below another in a matrix. When we now permute the 1st with the 3rd column and the 2nd with the 4th column there are still all possible permutations represented. However, now the 3rd and 4th sock (tuesday's socks) are chosen before the 1st and the 2nd sock (monday's socks). Thus, you can arbitrarily relabel the days.
Iku 😅 ini😊o7uh 8😮89i887ii8
At first, I thought the answer wouldn't be 1/9, because I thought the different scenarios would result in a different final answer. However, this math problem was a bit of an illusion for me. Here is how I thought through it.
Scenario #1: He selects a correct pair on day 1, so all remaining socks have a potential match for day 2.
There's a 1/9 chance that he chose the correct match on day 1. That means there are 8 socks remaining. The first selection on the second day won't matter, so there are seven socks left. There is a 1 in 7 chance that he makes a correct match. Between days one and two, there is a 1 in 63 chance that he gets them both correct.
Scenario #2: He makes an incorrect match on day 1.
On day one, there is an 8 in 9 probability that he made an incorrect match. This means of the 8 socks remaining, only 6 of them have a match remaining in the drawer. This time, the first sock he chooses does matter. It must be one of the socks with a matching pair remaining, so there is a 6 out of 8 probability. That leaves 7 socks remaining, and the one he chooses must match. There is a 1 out of 7 probability. That makes a 6 out of 56 probability (6/8*1/7) that the second day results in a correct match. Altogether, there is an 8 out of 9 probability that the first day was an incorrect match and a 6 out of 56 probability (or 3 out of 28) that second day was correct. Thus makes an overall probability of 2 out 21 (or 6 out of 63) that the first day was incorrect but the second day was correct.
Final answer: He could have a correct match on day two by being correct on day one (1 out of 63), and he could have a correct match on day two by being wrong on day 1 (6 out of 63). Together that makes 7 out of 63... otherwise known as 1 out of 9.
The answer of 1/9 is correct, and using formulas is much more efficient.
For me , this is the proper explanation. Thanks.👍
I wish I had an instructor like you.
I wish, I could have a teacher like you
be the same teacher for someone else
Like me?
i wish i had a dad
@@kolmaxikyes
On Monday he picks sock A and B. On Tuesday he picks the other sock from pair B. He now has 0 chance of getting a pair. The point is, your probability on days 2 onwards depend on what socks you picked on preceding days.
Also is all socks were the same, then he has a 100% probability.
This is really interesting to see. Ever since i've learnt nCr notation I've basically forgotten how regular probability questions work lol.
I got the same answer by just doing 5 x 1/10C2 which is 1/9.
In an Eddie Woo maths marathon now !
Same
ditto! haha
ⁿ⁷6😂😅@@angelstarfire
@@angelstarfirep are 😅8😢😂2
Such wholesome energy!
if a matching pair is taken day 1, then 4 pairs remain, so Prob of picking another pair is 1/7 ( Pr Pick a sock = 1 , then Pr picking matching sock is 1/7 )
if non -matching pair is taken day 1, 3 matching pairs remain and 2 odd socks.
Pr of picking either of those 2 odds socks Pr=2/8 - in which case it's impossible to find a pair
Pr of picking a sock that still has a pair in the drawer Pr=6/8
having picked a sock with a potential pair, picking that pair Pr = 1/7
Pr of picking a pair on day2 =
Pr picking a pair on day 1 ( 1/9) * 1/7
+
not picking a pair day 1 , (8/9) * picking a sock that still has a pair in drawing (6/8) * picking that pair (1/7)
=1/63 + 6/63
= 7/63
= 1/9 !
Why am I watching this for fun
Cause when we see someone enjoying what they are doing, we enjoy to see them, and math are beautifull too.
Eddie, please can you recommend a maths book I can use for my secondary education?
If I consider the matched pairs of socks (5) and the possible pairs (10*9), so I'll get p(E)=1/18.
Where am I doing wrong?
There are only 45 possible pairs, because pair AA and AA is the same (doesn't matter which A you picked first) - so possible pairs are 10*9 / 2. After that you get to 1/9 which is correct.
@@bl00dWILLfl0wyoh, thank you for the explanation 😅
If Monday has 1/9 of getting a pair, it's 8/9 it was not. If there is no replacement then Tuesday has 8 left with a 8/9 chance that two of them don't have a pair. The first sock's favorable set then becomes from 8 to six. So it is 6/8. Or 3/4. Then the second sock has a chance of 1/7 chance to match since there are 7 socks left. 3/4 x 1/7 = 0.10714285714 or 3/28. IDK but it seems like 3/28 would be it. Maybe I am overthinking this.
i didnt get from here
"so it is 6/8. Or 2/3"
how is it 2/3? isnt 6/8 = 3/4? and then answer should be 1/7 * 3/4 = 0.10714...
?
i thought the same!
the ending of video was rushed so i didnt get it
@@Ph0enix4862-H Dang. Nice catching that.
Omg I literally just did this exact thing before reading your comment. I do believe you’re right. Yes you would get 3/28 as the probability of getting a pair on Tuesday. But then, you’d have to multiply that by the probably of not getting a pair on Monday which was 8/9.
So 3/28 x 8/9 which equals 2/21.
@@caydenwaltersmusicYou then add that to the odds that the first day and second day were a match (1 out of 63) and you end up at 7/63 which is 1/9.
It seems a bit counterintuitive at first, but the math checks out. The probability for day 2 matches the probability for day 1 irregardless of whether day 1 was a correct match or not.
Does it mean that on Friday we have only one pair left and probability is still 1/9?
That's correct, because remember that probability is still tied to your choices from the previous 4 days.
@@emrecankarabacak oh yeah that makes sense. thanks (:
thank you so much, you really helped me with the correct point, hope you are fine🙂
I’m watching this for fun 😂
There are 5 distinct pairs of socks, each pair of a different color. Assuming all arrangements are equally probable. Suppose you pick two socks at random. There are in all C(10,2) = 45, combinations but there are only 5 which are favourable to the desired event which means the probability is .11111..... .
c(10,2) divided by 2 should be the total number of outcomes?
@@ashambarchaturvedi3344 No it is not. Not even probably and not according to probability theory.
@@sundareshvenugopal6575 gotcha
the ending felt rushed
I didnt get it!
Basically an answer that makes sense in theory but not in practice. I.E: Calculate the probability but don't assume that he's wearing the socks, he's just picking them.
I feel like it's misleading and would punish students who think rationnaly such as that kid who said "what about washing?"
bruh which uni is this i wanna join
When I draw randomly I never get matching pair. My socks are entangled socks.
Ideally there are 36 possible combinations in which 6 are perfectly matched. Hence the probability of getting a matched pair is 6/36 otherwise 1/6. So why not 1/6??
because you have to consider it not as picking two random socks at once, but picking one sock first and then another one.
Bill m l m l m l l ĺ l l l òl I ĺòĺĺĺĺm😊
Loop I
I wish I had u as my maths teacher!
2nd day, no 1st day match, 2/8 socks have you “drawing dead” after 1st draw. That feeling when you’re being taught how to think in terms of pokerr without realizing they’re being taught how to think in terms of poker….
Not sure if you are agreeing with the final outcome or not, but when you consider both scenarios for the first day (match vs no match), you end up with a 1 out of 9 probability for a 2nd day match.
I am afraid not. This is the probability of that desired outcome in exactly one trial of the experiment, not in seven trials of it. Week has nothing to do with it. In most finite probability problems you clearly identify the sample space of the possible outcomes in a single trial of the experiment and also those outcomes favourable to the desired event as a subset of the sample space and finally just take the ratio of their cardinalities. This is for simple events, for compound and complex events you will have to use other set theory as well as a bit more advanced probability theory concepts and principles. There must be a practical use of this theory for the advancement of human knowledge of the world we humans live in cause these can lead to a better understanding of the way this world truly scientifically works, with total perfection and precision, and also lead to a better appreciation of the underlying operational laws and principles governing the working of this world. It is certainly useful to us if we going to be in it. Or maybe we could simply sing, "We are the world". That may or may not cut it, depending on which tree you are talking about. Rather than simply say the quantum model is a non deterministic one for instance the seemingly random behavior of elementary particles could be seen as a probabilistic one, or in genes and vaccines the behavioral anamolies or abberations treated as probabilities. These principles can most certainly be grasped and applied in real life to solve real world problems. Although other than for my own personal use, with tremendous success, I have never really applied these principles in real world scenarios.
An honest question: If he only had enough socks for 5 days and not 7 sets of 5 days, would that change the probability in this case? I'm afraid he only has enough socks for one experiment. 😊 How do you see it?
As someone with basic mathematical abilities, I am impressed with the way you think. I seem to remember reading an article once about the effect that reiterations have on the probability of an outcome. Thanks for renewing my interest.
From which country is he??
This will probably not be answered, but my question is: If the question is without replacement, why is it that 1/9 applies on tuesday, wouldn't it be 1/7?
Because the probability of picking an acceptable 3rd sock is not necessarily 8/8 (1). Consider a case where on Monday you picked not matching pair. That means that on Tuesday the first sock you pick might not have a pair to pick anymore (as you picked it on Monday already). You need to consider both the case of picking a matching pair on Monday and picking non-matching pair on Monday and add the probabilities together.
Probability to pick a match on both days is 1/63. Probability to pick a match only on Tuesday is 6/63. If you add those together you get 7/63 = 1/9.
it was cruel of textbook authors ro leave out whether any socks are put back 😅
Shouldn't it be 8/8 * 1/7? on the next day! he said it is 1/9 no matter how you interpret it yet mondays socks are removed on tuesday...
New set of five everyday
Eddie bhai... Be safe in these Corona times. 😊
Its better to explain using Permutation and Combination
Vedio Quality ❌ Content Quality ✅
Boeing lost everything in macas and designs boring engineers cause you have only an ba maths licence of twarts converts ts if di falled hi teachers of college algebras .
Found this interesting
Hey could you write a little bit smaller please. It's just a bit large in my telescope viewfinder.
Your teachings is really great but did you build something new out of nowhere of course not because the complexes maths died not answer to the linearization of linears maths furthermore of complexes conundrums never stays the same as complexes systems cannot file the true nature of your self taught dusfuveried as a master can you explain why hiring list everything in your perfect linearisstion ion if slicing algebras algebras like a bra twarts if Boeing .
Its very easy i don't why your method is too difficult the simple answer is 5c2/10c2
No offense but maybe don't call it "very easy", call his method difficult, and then say the "simple answer" is 5c2/10c2 when 1) most courses learn probability before nCr notation and 2) you're not even correct it's just 5/10c2.
thank you very much
actually, i like the textbook explanation more.
😂
Milo
It's 100% bcuz the dude is a functional human being and uses his eyes for half a second to pick out a pair
китаец
I find the explanation unsatisfying, he probably elaborated on it more but the video ended before he could
Nice to find out he said the exact same thing after clicking on the next video 😂
Quit teaching
your name is rage, responding to a teachers video 8 years later lmao. you were in kindergarten 8 years ago