Worked Probability Questions (2 of 3: Selecting sock pairs)

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[@user-pv5hd1vu1t]

I like the way you used dominos to show the problem physically. I remember you teaching a similar style of question way earlier back, and i think this addition helps get a conceptual understanding of what's going on.

[@wizix9877]

are the socks discarded out of the drawer every day? because if they are put into laundry end of every day, then the last day probability would depends on the days before it, even the second and third day will depend on the previous day.

[@user-pv5hd1vu1t]

they are discarded. but even so. it is still 1/9. if there are 5 pairs of socks scattered so 10 socks and 10 people. each person goes to the drawer of socks and grabs a sock in turn w/o replacement. at the end they all reveal which sock they have. what is the probability that the first 2 ppl have matching pair. thats undeniably 1/9. but we can repeat a similar argument for persons 3 & 4, 5&6, 7&8, and 9&10 each corresponding to second third fourth and last pair respectively. because its without replacement. the people can all look away and someone else can just randomly rearrange the socks between ppl (or alternatively rearrange the people = rearranging the days). so any two socks that are matching pairs occuring next to each other out of all possible arrangements is probability 1/9. or two people next to each other having a matching pair is probability 1/9 which corresponds to each single day having prob 1/9 even without replacement i.e. even via discarding to laundry.

[@wizix9877]

@@user-pv5hd1vu1t ok I might be missing something. but the first day his chances of two similar socks is 1/9 so he may get those matched or he just lost two pairs all together. lets assume they dont match, second day there are only 3 pairs of matching socks in a total of 8 socks. so his chance of a matching pair second day should be 1/7 not 1/9 as the video explained. so first second and third days have different propabilities depending on the previous day if that person discards the sock out of the pool at the end of each day.

[@user-pv5hd1vu1t]

@@wizix9877 given that the first day didnt have a pair, yes there are only 3 pairs left for the second day. so choose one of the three pairs, (6 of the 8 socks) and then to match again, 7 socks remain so (6/8)(1/7) this happens 3/28 given first day doesnt match combining this, first day not matching is 8/9, so second day matching AND first day not is (8/9)(3/28) = (8/28)(3/9) = (2/7)(1/3) = 2/21 First and second day match is (1/9)(1/7) = 1/63 1/63 + 2/21 = 1/9 (second day match no matter what 1st happened on first day) If you do decide to go through all the hard calculations for all the nuances in 3rd/4th/5th day, it's all 1/9. I know it is counterintutive. Refer to one of my other comments that i replied to to another user.

[@user-pv5hd1vu1t]

@@wizix9877 If you really want the long way: doing it the long way. p(1st day match) = 1/9 p(2nd day match) = p(1st and second match or 2nd only match) = (1/9)(1/7) + (8/9)(6/8)(1/7) = 1/9 p(3rd day match) = p(all match) + p(second and third match only) + p(first and third match only) + p(third match only) = (1/9)(1/7)(1/5) + (8/9)(6/8)(1/7)(4/6)(1/5) + (1/9)(6/7)(4/6)(1/5) + ((8/9)(2/8)(1/7)(1/5) + (8/9)(2/8)(6/7)(4/6)(1/5) + (8/9)(6/8)(2/7)(4/6)(1/5) + (8/9)(6/8)(4/7)(2/6)(1/5)) = 1/315 + 4/315 + 4/315 + 2/315 + 8/315 + 8/315 + 8/315 = 1/9 p(4th match) = p(fourth and third match at same time) + p(4th match and 3rd doesnt) = (1/315)(1/3) + (8/315)(2/4)(1/3) + (2/315)(1/3) + (16/315)(2/4)(1/3) + (4/315)(2/4)(1/3) + (8/315)(15/2)(2/6)(1/5)(1/3) + (16/315)(15/2)(4/6)(2/5)(2/4)(1/3) + (8/9)(2/8)(1/7)(4/5)(2/4)(1/3) + (16/9)(2/8)(6/7)(2/6)(1/5)(1/3) + (32/9)(2/8)(6/7)(2/6)(4/5)(2/4)(1/3) + (8/9)(6/8)(4/7)(4/6)(3/5)(2/4)(1/3) = 1/9 ------------------------- Calculations: (1/315)(1/3) - all match (8/315)(2/4)(1/3) - two pairs out of contention (2/315)(1/3) - 1st and 2nd days only 2 colours removed, 3rd match (16/315)(2/4)(1/3) - 1st day no pair, 2nd day choose one of the colours from previous day and another colour not from first day (and viceversa) (4/315)(2/4)(1/3) - first, 2nd match, 3rd no match fourth match (8/315)(15/2)(2/6)(1/5)(1/3) - only one of the first two days match, 3rd no match (choose culprits from previous days), 4th match (16/315)(15/2)(4/6)(2/5)(2/4)(1/3) - only one of the first two days match, 3rd no match (choose one of the culprits from previous days the other pair is selected from ‘non-culprit’ and vice versa), 4th match (8/9)(2/8)(1/7)(4/5)(2/4)(1/3) - first day no pair, second day no pair because of first days socks (culprits), third day no pair, fourth day pair (16/9)(2/8)(6/7)(2/6)(1/5)(1/3) - first day no pair, second day choose 1 culprit and 1 non culprit, 3rd day choose culprits, fourth day pair (32/9)(2/8)(6/7)(2/6)(4/5)(2/4)(1/3) - first day no pair, second day choose 1 culprit and 1 non culprit, 3rd day choose one of the culprits only, fourth day pair (8/9)(6/8)(4/7)(4/6)(3/5)(2/4)(1/3) - first day no pair, 2nd day no pair from non culprits, third day no pair, fourth day pair --------------- Now let us look at 5th day p(all match) =(1/9)(1/7)(1/5)(1/3) p(two pairs out of contention, rest of pairs match) = (8/315)(2/4)(1/3) p(first two days only two colours missing, but not paired in those days, rest are pairs) = (2/315)(1/3) (16/315)(2/4)(1/3) - 1st day no pair, 2nd day choose one of the colours from previous day and another colour not from first day (and viceversa), 3rd day pair, 4th day not pair, 5th day pair (4/315)(2/4)(1/3) - first, 2nd match, 3rd no match fourth no match, fifth match (8/315)(15/2)(2/6)(1/5)(1/3) - only one of the first two days match, 3rd no match (choose culprits from previous days), 4th match, 5th match (16/315)(15/2)(4/6)(2/5)(2/4)(1/3) - only one of the first two days match, 3rd no match (choose one of the culprits from previous days the other pair is selected from ‘non-culprit’ and vice versa), 4th not match, 5th match (8/9)(2/8)(1/7)(4/5)(2/4)(1/3) - first day no pair, second day no pair because of first days socks (culprits), third day no pair, fourth day no pair, fifth day pair (16/9)(2/8)(6/7)(2/6)(1/5)(1/3) - first day no pair, second day choose 1 culprit and 1 non culprit, 3rd day choose culprits, fourth day pair, fifth day pair (32/9)(2/8)(6/7)(2/6)(4/5)(2/4)(1/3) - first day no pair, second day choose 1 culprit and 1 non culprit, 3rd day choose one of the culprits only, fourth day no pair, fifth day pair (8/9)(6/8)(4/7)(4/6)(3/5)(2/4)(1/3) - first day no pair, 2nd day no pair from non culprits, third day no pair, fourth day no pair, fifth day pair These calculations are the same as fourth day, as the fifth day is forced, it’s just that some days you need to swap fourth day to no pair for fifth day to force a pair. So p(last day = pair) = 1/9 Thus All are 1/9. Please, do not ever feel the need to do these calculations. It’s painful.

[@user-pv5hd1vu1t]

The other way to think about f) is to say that on any given day, we want said day to have a day where they chose a non-matching pair. If we use analogy of pairs = same colour, then two colours have been chosen out of the five colours. There are only three colours left, so only a maximum of three pairs of socks can be made. i.e. four pairs and one non pair is probability 0.

[@kking3875]

Great lesson 😊

[@that-plane-guy]

Great video!

[@philipmurphy2]

Great video

[@janis4909]

Nick picks two socks every day. how can the chance be 1/9 on the third day? If he picked two different socks on the first day and two different socks on the second day there would be only 2 matching socks out of 6 left?

[@Acharya_Slideshare]

Now this is mathmatication of my brain wow really enjoyed the question

[@Ynook]

I am not sure if I got it. Each day is independent of each other, isn't it? On the first day, I have 10 options, on the second day I have 8 and so on. How can the probability be 1/9 for each? Why isn't it 1/9, then 1/7, then 1/5 and so on? Of course, that's presuming the pair of the first sock I draw is still left in the drawer. If, for example, I draw blue and yellow on the first day, then if I draw yellow on the second day, the probability of making a pair is 0. So, starting day 2, there should be two cases considered: one if the sock I draw first still has a pair left, one if the sock I draw has no more pair left. Of course, in the real life, I can wear no socks if I don't draw a pair and I can keep my socks for the next day. If I don't draw a pair on the next day, I can pair them with what I got from the day before. This way it would make sense why the probability is 1/9, but that's not what we're calculating.

[@user-pv5hd1vu1t]

For the sake of simplicity, I will only describe the case where we just go one further, because it gets messy afterwards. Second day = pair, what is the probability of that occurring? Case 1: First day is a pair This happens 1/9 8 pairs left. So second pair is 1/7 So 1/63 for both days. Case 2: First day is not a pair, e.g. like ur case blue and yellow This happens with probability 8/9 So on second day there are 8 socks left But two of those socks (blue and yellow) are single loner socks So to create a pair on the second sock, we only have 6 valid socks left out of the eight So the first sock on second day must be made with probability 6/8 = 3/4 Then the last sock to make pair on 2nd day is 1/7 So 2nd pair is (3/4)(1/7) = 3/28 So first day not pair + 2nd day pair = (3/28)(8/9) = (2/21) = 6/63 Case 1 + case 2 = 1/63 + 6/63 = 7/63 = 1/9 ----------------- We can only use the 1/9, then 1/7, then 1/5 and so on given that the previous days had pairs too. so p(2nd day pair | 1st day pair) = 1/7 since there are 8 socks left and 4 pairs to choose from p(3rd day pair | previous 2 days paired) = 1/5 6 socks left, all pairs still. no weirdo situation. So what ur thinking about is the conditional probability. not p(day x is a pair) in general. ---------------------- Of course, in the real life, I can wear no socks if I don't draw a pair and I can keep my socks for the next day. If I don't draw a pair on the next day, I can pair them with what I got from the day before. This way it would make sense why the probability is 1/9, but that's not what we're calculating. (If u think about it carefully, this is actually what the question is implying. i know its counterintuitive). I try to wrap my head around it by thinking of a simple-ish game. There are three eggs. 1 is raw and 2 are hard boiled. There are 3 players. Each player takes turns to choose an egg, then at the same time they crack it on their forehead. If I am one of those players, what is the probability I choose the raw egg? Does it matter if I choose first , second or last? Order doesn't matter, still 1/3 all the way through. So same analogy with 1/9. Even though the calculations are picking day x socks on day 1, this is actually valid through said idea from the eggs situation.

[@Ynook]

@@user-pv5hd1vu1t I think your explanation made me understand it, especially the egg part. Thank you.

[@hanaelkamhawy7194]

ABC is a right-angled triangle at B where 3AC = 5BC , then tan A =3/4 Can someone please explain why tan A =3/4

[@kevinsessoms1982]

I really like this guy! He's really good at explaining math concepts. I do disagree with him on the fact that knowledge of the previous days choices do not affect the probability of the last day's choice. If you knew that the person chose perfectly matching colors on the previous 4 days then the probability is 100% that the last day the colors will match. There is not enough information to actually answer part of the question. There is also the probability that the previous 4 days the colors were not matching but 4 matching colors were used up leaving on one color.

[@aussie807]

ahh, but what if he put the dirty socks in the wash each day🧐🧐

[@user-pv5hd1vu1t]

thats what the whole question is implying. (socks to last the whole working week). so it implies they wear a different pair each morning. for a) b) c) the answer is still 1/9 whether u replace the socks or w/o replacement where this question implies without replacement is going on. its counterintuitive i know. if you consider all possible permutations of these five pairs of socks. define the first two positions containing socks to be the first day socks, the next two for the next day and so on... there are 5 pairs in total so 10 socks which is 10! but there are 5 pairs of repetition of 2! so total amount of permutations is 10!/(2!)^5 if we fix any two positions next to each other (some representing days of the week) and restrict this two having a pair of socks, the number of permutations for this restriction is 5×8!/(2!)^4 since there are 5 total matching pairs that can go in the restricted two slots and we still gotta permute the remaining 8 so the probability of this happening is (5×8!/(2!)^4)/(10!/(2^5)) = 10/90 = 1/9 so according to this p(first two socks is a match = first day's socks) = 1/9 p(2nd sock in first day and 3rd sock for 2nd day is match) = 1/9 p(3rd and 4th sock match = 2nd day match) = 1/9 ... p(ith sock and (i+1)st sock match) = 1/9 for all i = 1, 2, 3, ... , 9 this makes 9×1/9 = 1 covering 'all cases'. so for any morning, even through discarding into the wash or whatnot is 1/9 ------ if u wish to see other explanations. refer to my other replies to comments of other users in this video

[@user-pv5hd1vu1t]

if it didnt imply that the socks go into wash... then the answers for d e and f would be different d) (1/9)^2 = 1/81 with replacement (can use same socks from previous days) 1/9 × 1/7 = 1/63 (they go in the wash) e) (1/9)^5 (can use same socks, with replacement) 1/9 × 1/7 × 1/5 × 1/3 (without replacement, socks go in wash) f) 5 × (1/9)^4 × (8/9) (when a pair is chosen. put it back in drawer so subsequent days still have 5 pairs to choose from. there are 5 days to choose from to decide its a non match day.) 0 (impossible to happen when they go into the wash. max three days matching of one day not match)

[@aussie807]

Thank you but I meant as whimsical reply, not literal. But it highlights to only read the question literally. Many students will try to read implied information into question that can get them into trouble.🙂@@user-pv5hd1vu1t