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.
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.
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….
But what if the first day A and B is chosen. That means that if A or B is picked the second day. It is impossible to get a matching pair. As they have already been picked.
Yes but if the first day both A socks are picked, than you have a probability of 1 × 1/7 = 1/7 to pick a maching pair on day 2. The best way of proving that P ( match 2nd morning) = 1/9 is with a probability tree but it's a bit long
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..... .
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.
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.
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.
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?"
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 !
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??
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 .
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
In an Eddie Woo maths marathon now !
Same
ditto! haha
ⁿ⁷6😂😅@@angelstarfire
@@angelstarfirep are 😅8😢😂2
Eddie, please can you recommend a maths book I can use for my secondary education?
thank you so much, you really helped me with the correct point, hope you are fine🙂
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
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.
Such wholesome energy!
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….
When I draw randomly I never get matching pair. My socks are entangled socks.
But what if the first day A and B is chosen. That means that if A or B is picked the second day. It is impossible to get a matching pair. As they have already been picked.
Yes but if the first day both A socks are picked, than you have a probability of 1 × 1/7 = 1/7 to pick a maching pair on day 2.
The best way of proving that P ( match 2nd morning) = 1/9 is with a probability tree but it's a bit long
Hey could you write a little bit smaller please. It's just a bit large in my telescope viewfinder.
Watching math for fun now !
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
Eddie bhai... Be safe in these Corona times. 😊
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 (:
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.
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
@@humamubarak6411 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.
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.
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.
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?"
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 !
it was cruel of textbook authors ro leave out whether any socks are put back 😅
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
Found this interesting
thank you very much
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
actually, i like the textbook explanation more.
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 .
😂
Its very easy i don't why your method is too difficult the simple answer is 5c2/10c2
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 .
китаец
It's 100% bcuz the dude is a functional human being and uses his eyes for half a second to pick out a pair
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