A legendary question from the toughest exam

[@user-jn4sw3iw4h]

With on average 3 minutes per question, through a 6 hour gauntlet the correct answer to this question is "skip, lets see if there will be time left to loop back around"

[@mj9765]

Most other questions are the same way. The students who answer these have a muscle memory of solving such questions for years, sometimes, unfortunately for a decade! Yes, students start at 8 or 9 yrs of age to prepare for this exam which they take at 18! Very sad!

[@user-jn4sw3iw4h]

​@@mj9765 even with "target audience, is a lot more fluent with these terms, as they prepare for this exam for nearly a decade" doesn't change the argument. In a "3 minutes per question, (2x) 3 hour gauntlet"-exam, tactically picking which to skip, is undeniably part of what's being tested. For such an established institution, I'm going to assume, this is intentional. Given the "less than 0.05% succes-rate, even among the target audience of the exam" it is statistically like, this was one, you were better off skipping. My statement was independent, of the specific contents of the question.

[@kaushikmohanta8526]

You are right. It's a game of rejecting these questions where the answer would take up at least 6 min of your time Rather solve the easier ones first. And if you feel like you can score this one, then only go ahead. For a tough exam like this, the cutoff usually goes low personally, I'm skipping this one!

[@lamelauncelot1385]

Don't make stuff up​@@mj9765

[@yashagrahari]

@@mj9765 They enjoy solving these problems. So, there is nothing sad or unfortunate.

[@ai314159]

What makes this question hard is not the math-it's the incredibly confusing wording.

[@robertveith6383]

* confusing Use the adjective.

[@lefthanded3512]

Haha, exactly. The problem itself is actually extremely easy if you think about it logically and methodically. The way it’s worded makes it sound like a question only someone with an IQ of 180+ can answer correctly.

[@pierreabbat6157]

The way I understood the question on reading it, I got the answer 6*57+15*42+20*22+15*7+6*1=1523, where 57 is 15+20+15+6+1.

[@ai314159]

@@pierreabbat6157 same here

[@ayushrudra8600]

yeah i solved it pretty easily with contest math background - it was probably just a combination of probability not really being taught in school and the wording

[@StatsJedi]

At the pause, the definition of (A,B) is totally unclear. No wonder so few got it correct.

[@chuckywang]

Exactly! I thought A and B were just two independent subsets of S.

[@sonure6127]

What is so unclear in the definition?

[@TomNimitz]

@@sonure6127 The concept of "independent" in the solution is flawed in the solution, where the poster has applied some twisted logic that actually applies a dependency between A and B that excludes certain otherwise-independent (A, B) combinations separate from the explicit 1

[@sonure6127]

@@TomNimitz Events X and Y are independent means that P(X intersection Y) =P(X)P(Y) . I hope u get it now.

[@sakumar]

An event is defined as a subset of the sample space. So, for example, you could say, "If I toss the dice, what is the chances it comes out

[@AdrianColley]

I would have skipped this question because its phenomenal vagueness is not worth spending exam time on decoding. A must be a set, because |A| is defined only on sets, but it's also an "event associated with S"; which can't indicate the natural meaning of an element of S, because A's cardinality is greater than 1. Whoever wrote that question should have had their work reviewed before it went to print.

[@chanlaoshi8634]

An event is defined to be a set in probability theory, so the formulation is ok for people who know this term. But when writing a text I would not expect every student to know that and be a bit more clear.

[@harshuldesai8901]

Well the kids here (India) learn that an event is a subset of the sample space. I honestly didn't understand what the confusion was about but maybe I'm starting to realize the definitions taught aren't the same everywhere. But I do believe this question made sense for most test takers.

[@AdrianColley]

@@harshuldesai8901 I think you've nailed it here. I was taught probability theory at both second and third level in Ireland, and I never heard that term "event" used with that meaning.

[@anibeto7]

​@@AdrianColley Definition of an "event" is an outcome or a set of outcomes from a given experiment.

[@go_gazelle]

My issue is with independence vs dependence. I interpreted "independent" to mean the probability of B does not depend on A (and vice versa), meaning I'm just picking two sets at random. 63 possible sets for A and the same 63 for B. (Null set excluded). 63² total possibilities for (A, B). (6 + 15 + 20 + 15 + 6 + 1 = 63) or (2⁶ - 1 = 63) 923 (A, B) for which |A| = |B|. (6² + 15² + 20² + 15² + 6² + 1² = 923) 63² - 923 (A, B) for which |A| ≠ |B|. (63² - 923) / 2 (A, B) for which |B| < |A|. Put that in a calculator: 1523 (A, B). As well, this seems reasonable to be completed in 3 minutes.

[@sergniko]

I paused video to understand the question

[@strangelyrepulsive77]

i paused to video to wonder at that i will never even understand it

[@mrbutish]

😂😂😂😂

[@pablocopello3592]

A mathematical problem begins when it is clear what is being asked. Before that, we have a linguistic/notation/convention problem like in this case. Does JEE tests evaluate the knowledge/capacity in it's target topics, or evaluate the capacity to guess what it's questions pretend to ask?

[@astrogersunited5389]

the competition is high hence such questions are asked

[@ParitoshTripathiOfficial]

The director of exam board himself said we conduct this exam to eliminate students and not select them. They have to remove students by hook or crook. So, they resort to such techniques by making question ambiguous, incorrect, having multiple correct answers. That is why JEE have the reputation of toughest because no one can guess all questions correctly.

[@ParitoshTripathiOfficial]

The irony!

[@Prabhu108.]

Very interesting point. Even a smart person would be troubled if they aren't familiar with the wording.

[@chair7728]

Obviously math is taught differently around the world but when I read the question I found the wording to be very mathematically exact and completely unambiguous, with the terms having a strict mathematical definition

[@fun-damentals6354]

i feel like the problem itself is actually rather simple. its the fancy mathematical terms and confusing wording that makes it difficult to understand

[@asdfasdfasdf1218]

The terms themselves aren't exactly fancy, it's just that they are the strict and formal definitions that nobody uses in everyday speech.

[@lefthanded3512]

You find the true smart folks by asking hard questions in an easy format rather than by asking easy questions in a hard format 👍🏻

[@Kounomura]

I don't quite agree with that. Because it is also a valuable skill that someone can peel out of the confusion what it is actually about.

[@lefthanded3512]

@@Kounomura I can agree with that to an extent. It depends what you’re looking for, a linguist, a problem solver, or both.

[@prometheus7387]

​@@Kounomura that's for the English exam to test IMO, not the Math exam

[@thechessplayer8328]

It seems to me that in science, one tries to explain something that no one understands in a way that everyone can understand, and in poetry, it's the exact opposite.

[@tinnderbox3410]

Very much agree here. This problem is at least partly a language question. I'm fairly familiar with probability theory but I didn't understand this problem at all. For those who had studied with a different textbook (calling a subset a 'subset', for instance) this problem was impossible. Small wonder almost no one got it right.

[@GaurangAgrawal2]

I started watching your channel in 8th grade and I was always amazed by how you solved JEE Problems. Especially the ones involving logarithms. Ofc, I wasn't able to understand anything at that time, but now that I am preparing for JEE, It's time to correctly solve this problem in order to check my preparation 😅. Loved these years with you and your way teaching mathematical concepts. Looking for more in the future Edit: Nah This question was from my least favourite topic Permutations, Combination and Probability... I am currently re-studying those topics to understand them better.

[@deadlock7946]

I wish you the best! Never stop solving and enjoy the process❤

[@egaLwie.]

wish you best of luck

[@makokx7063]

Imagine getting through that whole thing then messing up the final addition. lol

[@arthatiara4091]

what im thinking when presh said there were only 78 students, "did they have enough time to finish other questions???"😂

[@OrbitTheSun]

Maybe they just got lucky with their guess. Can someone calculate the probability?

[@howareyou4400]

This took me about 2 and half minutes to solve. But I took probability class in undergraduate study. I wouldn't be able to understand the concept of "event" there back in high school as it wasn't taught.

[@afzal_amanullah]

​@@howareyou4400What do you mean by you wouldn't understand the concept of Event in your highschool Isn't probability taught in highschool?

[@howareyou4400]

@@afzal_amanullahBasic probability, yes. But the concept of "event", the one that utilizes the set theory isn't taught until university. You can see that many people in the comment section complain that the "problem description isn't clear". That's likely because their probability class didn't taught the "event" as in the set theory.

[@afzal_amanullah]

@@howareyou4400 what country are you in?

[@aliso-pv7ll]

“Oh is a letter, zero is a number!”

[@vincent412l7]

Nought is a number

[@ravirajshelar250]

The 2019 paper is always going to be on the top 5 hardest papers in JEE Advanced History😂. Also in jee advanced papers students generally leave questions related to permuations and combinations a lot as compared to other topics.

[@anuragiyer4870]

PnC was my favourite topic and I made sure to do it first always

[@Kounomura]

The strict mathematical language can be so abstract, even in the formulation of the simplest problems, that it directly scares off those who want to get to know the beauties of mathematics. Above all, in education, much more care should be taken to make every mathematical problem as close as possible to an imaginable situation. Once the listeners have understood the essence of the matter through this, it is then possible to continue with the more abstract and generalized version. In the case of the above test, it is not possible to know exactly who they wanted to select in the first place. Those who switch quickly or who have a large overview, even if it is not that deep.

[@olerask2457]

I guessed 1523. I read the sample space as rolling a dice, and the independent events as rolling TWO dice. I then had to calculate the sum of all products of binomial coefficients (6,a)*(6,b) under the condition 1

[@chiragmidha2878]

I am glad that the Jee exam is recognized as one of the touhgest exams in the world. I gave the test in 2017 and because of that I am still able to follow these math videos, even though I haven't studied math in a while now. Also regarding the percentage of people who got the correct answer. Population of India is huge so a big majority of test takers don't really prepare to that level, however to even give the Jee advance test, you have to get more than certain rank in Jee Mains (an easier version test, (you only have 2 minutes per question in this exam but the questions are easier)). The point that I am trying to make is people who even attempt the Jee advance test have already gotten really scores in the main test.

[@MoonGlow22]

I tried to solve in a diffrent way and find the answer 1523 I guess I didnt considered that they have to be independent and did the 10:30 calculation for 6 times İf |A|= 2 3 4 5 6 Then --------------- |B|=1 2 3 4 5 1 2 3 4 1 2 3 1 2 1 Then we calculate Combinations for each one which leads to 15 20 15 6 1 ------------------------ 6 15 20 15 6 6 15 20 15 6 15 20 6 15 6 İf we sum all possible B values for each column we get 6 21 41 56 62 Then 15x6+20x21+15x41+6x56+1x62 = 1523

[@MoonGlow22]

Also its 23:21 now and I have a electromagnetic wave theory exam tomorrow 😅

[@SLuce222]

I got the same answer (1523) different method.

[@samarthpawar1504]

i got the same answer...and i directly skipped to the end to see if i am right and then comes disappointment😮‍💨

[@trilokvyas7987]

in the same boat

[@muthiah9705]

I am an Indian student and preparing for this one . Wish good luck for me.

[@brahmbandyopadhyay]

Good luck ~ from WB

[@melanieelayaperuma3110]

Google luck, tell us your results!

[@lefthanded3512]

Good luck man 💯 you got this

[@ancientancedotes]

@@brahmbandyopadhyay me too........i am a future aspirant

[@KrishnaThakur-nf4rt]

How was ur advanced?

[@leogreaves3251]

It’s always a treat when I get an upload notification from you.

[@dhananjaysawant4646]

1:14 for a second I thought that was the math problem

[@GWF_YOU]

🤦

[@JaneAustenAteMyCat]

It'd be a nice, easy question if it were!

[@davidmagdalena8149]

🤣🤣🤣🤣🤣🤣🤣🤣

[@Vishal00567]

I also thought same.

[@dannyannet154]

😭🤣🤣🤣

[@Gyan-fx9zx]

Thanks for making this video on my suggestion.

[@anishgain2563]

From the title i knew it was gonna be a question from jee advanced

[@aayushprasad7318]

Our Jee advanced 2024 exam after 7 days. Wishing all the best to all jee adv 2024 aspirant !

[@Shubhu519]

how did it go?

[@aayushprasad7318]

@@Shubhu519 scored 104. Cutoff was 109. Not selected.

[@KrishnaRajoriya-su2di]

I got 107 marks cutoff was 109 marks when I returned home and attempted the paper again I was scoring 150 marks because I did a lot of blunders during the exam Feeling very disappointed as I only scored 95 percentile in JEE Mains I could have easily cracked 2024 advanced paper but did a lot of mistakes there were so many questions that I could have solved easily but failed due to pressure of exam Thinking of taking a drop

[@megabot49]

Thats legendary man

[@HikingEngineer]

yeah this question is bogus -- and i have a math degree -- i'm assuming everyone put 15 as the answer?

[@bobh6728]

I vaguely remember this from Probability and Statistics class years ago. For example, using dice the sample space S is {1,2,3,4,5,6}. Then you roll the dice (die) once and get a result. That is the experiment. They you can enumerated all the events. An event would be like rolling an even number (2,4,6), or rolling a prime (2,3,5,7). All the events would be the power set of S. Null set (1), (2), (3), etc (1,2), (1,3), etc until you have all subsets. So the wording is standard for sample space, events, independent events, and cardinality.

[@robertlezama1958]

Agreed. How does an event become a set of elements? The dice reference in the explanation seems to establish that the displayed value is independent of the other 5 possible outcomes, without using probability. Ordered pairs of events where the value of 'A' must always be > that of 'B' led me to 15 for the answer. Or maybe I should be back in math classes. 😢

[@bobh6728]

@@robertlezama1958 The events are sets of elements. Use rolling a die. There are six possible outcomes for the experiment. But there are 64 events that can be described which are the power set. For example, one event would be throwing an odd number {1,3,5,7} or an even number {2,4,6,8} or less than 4 {1,2,3}. There are 64 different events, which are all the subsets of {1,2,3,4,5,6}. Many are difficult to describe is words like {1,3,4,6}, but it is an event.

[@asdfasdfasdf1218]

@@robertlezama1958 An event is *defined* as any subset of the sample space. If you don't know the formal definitions of all the terms, then obviously the problem is hard.

[@anshumanagrawal346]

How do you have a math degree and not get this? I can understand for a highschooler to not get it for once; but if you have ever formally studied any probability, you should have no problem with the idea the Events ARE Sets, and that's what probability is assigned to, and then just the definition of independent events.

[@JEE-oq1me]

thank you, this is just what i wanted to see 12 days before giving JEE Advanced myself....

[@Kshitij.with.nature-channel]

😂😂😂

[@hamburgeryumyum7491]

How did it go?

[@JEEsiast]

Bro what was ur mains percentile???? Btw how did ur advanced go? mine was good😌

[@Szynkaa]

it's easy and clearly worded (unlike some people in comments suggest), however 3 minutes for this question is incredibly harsh timer

[@verkuilb]

@9:35, you say that with |A| = 3, the only way to get a multiple of 6 is if |B| = 2. Not true-you can also get a multiple of 6 if |B| = 4. I realize that this would violate the rule that |B| < |A|, which I suspect is why you skipped it, but I believe you forgot to state that.

[@mk2k685]

paused right there and skimmed the comments, thanks for clarifying this

[@DrHenry1987]

It makes you wonder if the problem had appeared elsewhere and those 78 had read the answer and remembered.

[@HikingEngineer]

agreed

[@BuddyReiner]

Nah, each year JEE advanced questions are made totally new, with rare exceptions they give modified and simplified IMO problems sometimes.

[@anshumanagrawal346]

I don't know why you say that. I saw this probably for the first time when I saw the video and was able to solve it in a couple of minutes. It's been 2 years since I graduated high school, so I don't exactly remember it fresh, if you know the definitions and are comfortable with counting (combinatorics/PnC) then there really isn't much.

[@DrHenry1987]

@@anshumanagrawal346 Good for you.

[@omm-o6i]

​@@anshumanagrawal346👏👏👏

[@drelijahmikail3916]

A \intersect B can be 0, e.g. B = {1}, A = {2,3,4,5,6}, which satisfy 1

[@gameobee777]

But it doesn't satisfy the "independent event" condition. I.e pr(a intersection b) = pr(a)pr(b) Because if a intersection b=phi Then, Pr(a intersection b)= undefined 😊

[@mwv1969]

I thought that I completely misread this as the equivalent of rolling two dice (A, B) and needing to identify the number of pairs where B is between 1 and 5 where B is < A (so B/A with the pairs being 1/2, 1/3, 1/4. 1/5, 1/6, 2/3. 2/4...4/5, 4/6, 5/6) giving me 15 ordered pairs. I drop this problem into ChatGPT asking for the solution and it also comes up with 15 as the answer. Here is the cut and paste solution from ChatGPT: Let's break down the problem step by step: We are given a set (S = { 1, 2, 3, 4, 5, 6 } and we need to find the number of ordered pairs ( (A, B) such that 1

[@salihkizildag]

it's not a pair of numbers, it's a pair of events

[@OceanusHelios]

This problem is difficult not because of the arithmatic involved. This problem is difficult because it is requires knowledge and familiarity of a subset of mathematics that is rarely called into use, and so the level of familiarity with it will be very low. Performance will therefore be based more on how exhaustive a person's education in math is with regards to the landscape of maths overall, than the counter-intuitiveness or complexity of the problem. In other words, if a person was used to solving this type of problem and answering these kinds of questions, it would have been a no-brainer. Just as with the rest of Chemistry and Physics included in this test, it will be a test of the thoroughness of the person's education and a familiarity with the material. It isn't a test of intelligence or anything of the like but a test of skill and how comprehensive the person's knowledge is. In giving a test with these kinds of problems, they could extend the time of the exam to three weeks and would students would still have roughly the same scores. If the exam is meant to weed out and select for the "best of the best" then it is a good test to find your super-nerd in the bunch of test takers.

[@captainoates7236]

I naively read the question to mean if B is 1 or more and A is bigger than B then the ordered pairs available would be: 1,2 1,3 1,4 1,5 1,6 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 and 5,6. In other words 15 pairs.

[@HenrikoMagnifico]

Ok how the heck can you do this in 3 minutes

[@sparky2141]

That's the thing You don't :) You solve the other "easier" questions in less time and invest that saved time into this question Hence the exam also tests your time management skills

[@Ibrahim208]

you either skip the question (You can do that in this test) and have extra three minutes for other easy questions. Or you sacrifice some easy questions to attend this question (there is a chance you may get this wrong).

[@fade2dblack]

Question starts at 1:47 if you want to skip the visualization of a small number not relevant to the question.

[@imeprezime1285]

Hasebauer-Dickheiser Test, problem #14 baby!

[@Bdple]

Its a quite easy problem for any indian student who is seriously preparing for JEE Here's the solution. 1

[@Bdple]

Anyone who is speaking about the language of the question.. Let me tell you this question is nothing just take any jee mains or adv paper you will get to know.. There are many many many questions with more complexity..They put sets and permutations and combinations in any question, making a good hybrid..Students are used to it...

[@Bdple]

It's more or less a counting question.. They want students to waste their time thinking. Makes it a trap..

[@deepvaghasiya3597]

The question is not ill written, it makes sense and it is written perfectly fine. The thing is, the formal definitions are unclear to many people, and so to many students preparing for JEE. In India, students prepare for JEE from coaching institutes, and they usually don't teach formally, they just teach enough to crack the exam, that is they just give mechanical definitions and ideas just to solve normal questions, it is those few students who actually read good books and learn the formal definitions by themselves. Thats the reason this question wasn't solved by many students, whereas it was just the use of basic definitions thats it, the question was lengthy so tough to solve it in given time bound, but conceptually, it was not that tough, just the use of basic definitions.

[@Tiqerboy]

The answer approaches infinity. The problem with this problem is that it doesn't put any bounds on A and B. The set S suggests a six sided die.. I'm interpreting events A and B representing an arbitrary number of rolls of the die. I could roll the die as many times as I want and still have an event A that has as many elements as I want. Therefore if B has just ONE element (i.e. one roll of the die) then there can be an arbitrary number of ordered pairs (A,B) where A is as large as one likes.

[@harshuldesai8901]

If we assemble a deck of 52 playing cards with no jokers, and draw a single card from the deck, then the sample space is a 52-element set, as each card is a possible outcome. An event, however, is any subset of the sample space, including any singleton set (an elementary event), the empty set (an impossible event, with probability zero) and the sample space itself (a certain event, with probability one). Other events are proper subsets of the sample space that contain multiple elements. So, for example, potential events: "Red and black at the same time without being a joker" (0 elements), "The 5 of Hearts" (1 element), "A King" (4 elements), "A Face card" (12 elements), "A Spade" (13 elements), "A Face card or a red suit" (32 elements), "A card" (52 elements). This is from Wikipedia. Maybe you need to revise your definitions. An event is a subset of the sample space which is S here. Even after an arbitrary number of rolls, the results will always be a subset of S and so the cardinality of both A and B will always be less than or equal to 6.

[@hhhhhh0175]

this question doesn't make any sense unless S is supposed to be the power set of {1,...6}.. how is the "probability" of a subset equal to its number of elements divided by the number of possible elements? that's stated nowhere in the question

[@deepvaghasiya3597]

Are you a JEE student? Because they don't teach the proper definition of probability, they just say 'favourable outcomes / total outcomes' So, here's the actual definition of probability (discrete probability to be exact) probability is a function P : S -> [0,1], where S is the sample space, and for discrete sample spaces let P_w denote the elementary probability of element w in S, which is 1/|S| here as all are equally likely (probability of any single element), then probability of an event, which is a subset of the sample space, say A, is given by P(A) = Summation(P_w) for all w in A, hence, it will be |A|/|S|.

[@kilimanjarocruz660]

I totally agree. This is ill-defined in this way.

[@jonathanlerner2797]

This is exactly what threw me off. The proper probability of any particular subset of S is (1/2)^|S|.

[@hhhhhh0175]

​@@deepvaghasiya3597 i've never seen that definition before. is it supposed to correspond to normal probability for single element sets, and then "extend" it for any subset of the space? is this a well known thing in the context of the exam? and what does it have to do with probability? it seems like a pretty arbitrary definition to me, but i don't live in india, so maybe i'm missing something

[@lefthanded3512]

They probably assume you will deduce that yourself. If there’s a logical way to do that, then I can see them testing you that way.

[@powergi3996]

The best thing to do on a question like this in an exam is to skip it and use the time on other easier questions, especially if all the questions are worth the same. I'd be very curious to know how much time these 78 people used to answer this.

[@_Dearex_]

even after the explanation i have no idea what this questions wants from me

[@anibeto7]

It wants your blood.

[@bobtoad8601]

The difficulty is how badly the question is written.

[@Shunyasium]

I appeared for JEE advanced 2024 yesterday!

[@musicmaker99]

I don't understand the question, i don't understand the answer, and i don't understand the explanation. 422 what? How does this relate to dice-throwing?

[@Ibrahim208]

So basically, two people are rolling a dice |A| and |B| times respectively (Here |x| doesn't mean Absolute value of x, it represents the number of elements in the event set X ie Number of times the dice has been rolled) .The question is asking how many such possibilities of A and B exist such that it satisfies 1

[@kapilchhabria1727]

I took the IITJEE in 1999 and had a rank of 5000, which is not high enough to secure admission into the program of my choice. When I took it was a 3 hour mathematics and 3 hour physics exam on day 1 and 3 hour chemistry exam on day 2. The exam had two sections, MCQ of which more than one choice could be correct and that was followed by long form questions. Additionally you are not allowed the use of a calculator nor are you permitted to use logarithmic tables. This meant deriving square root using the long form method.

[@nickronca1562]

Without watching the video, I almost got tricked by the wording, but after going over what I think the wording means, I got 1,523.

[@MichaelRothwell1]

It took me while to get what the question was about, and another while to figure out a reasonable strategy, but in the end my solution was almost identical to the one in the video. 1≤|B|0, |A∩B|>0. If |A|=2, then |B|=1, |A∩B|=1, so 6|A∩B|=6 and |A||B|=2, so 6|A∩B|=|A||B| is false. If |A|=3, then |B|≤2. 2|A∩B|=|B|, so 2 | |B|, so |B|=2, |A∩B|=1. 6C3=20 possibilities for A, 3C1×3C1=3×3=9 possibilities for B in each case, so 20×9=180 possibilities. If |A|=4, then |B|≤3. 3|A∩B|=2|B|, so 3 | |B|, so |B|=3, |A∩B|=2. 6C4=6C2=15 possibilities for A, 4C2×2C1=6×2=12 possibilities for B, so 15×12=180 possibilities. If |A|=5, then |B|≤4, |A∩B|≤4. 6|A∩B|=5|B|, impossible as 5 does not divide into 6 or |A∩B|. If |A|=6, then |A∩B|=|B| (automatically true anyway). 1 possibility for A, B can be any subset of A of size 1 to 5, i.e. 2⁶-2=64-2=62. So 180+180+62=422 possibilities altogether.

[@Player_is_I]

Hats off to "THE 78"

[@TheRMeerkerk]

After googling what half these words mean and trying for smaller sample spaces I came to this: There are only going to be independent events for |A|=6 and 1

[@TheRMeerkerk]

If |A|=4 and |B| =|3|, then P(A)P(B)=1/3. For P(A intersect B)=1/3 to be true, B must have exactly two elements in common with A. There (4 choose 2) ways to choose those elements and (2 choose 1) way to choose the third element which totals to (4 choose 2) x (2 choose 1) = 6 x 2 = 12 possibilities. However, we must not forget that there are (6 choose 4) ways to choose elements for A, which means 12 x (6 choose 4) = 12 x 15 = 180 ways to choose |A|=4 and |B|=3 such that they are independent.

[@TheRMeerkerk]

If |A|=3 and |B| =|2|, then P(A)P(B)=1/6. For P(A intersect B)=1/6 to be true, B must have exactly one element in common with A. There are (3 choose 1) ways to choose that element and (3 choose 1) ways to choose the other element of B, which totals to (3 choose 1) x (3 choose 1) = 3 x 3 = 9 possibilities to choose B. However, we must not forget that there are (6 choose 3) ways to choose elements for A, which means 9 x (6 choose 3) = 9 x 20 = 180 ways to choose |A|=3 and |B|=2 such that they are independent.

[@TheRMeerkerk]

If we add all of them together we get a total of 62 + 180 + 180 = 422 pairs (A, B) such that 1

[@aroundandround]

4:30 What does the event A mean to result in 1/2 probability for a set with three of the six elements?

[@user-bg2ct4dy5k]

You failed be cause you took 15 minutes to solve a 3 minute problem.

[@mr.d8747]

*I wanted to say that first:(*

[@Schrodinger0]

He is explaining the problem, not solving it.

[@Boolarramob36]

Mean of 3 minutes not mandate of 3 minutes. Made me laugh though.

[@jkid1134]

I mean, he also probably just looked it up, which in a test environment would not fly either.

[@Maths_3.1415]

​@@Schrodinger0 These guys just want to show off that they're something special.

[@navghtivs]

I think the hard part is to understand the wording, once that's understood, I'd just write a python program to enumerate all combinations and get the answer in no time.

[@asdfasdfasdf1218]

Probability theory has a knack for formally defining many terms that are different from ordinary usage in English.

[@anshumanagrawal346]

Lol

[@deborahd.7281]

It took about 13 minutes in the video to solve it and the students had on average 3 minutes to solve it.

[@randomguy4803]

😂

[@bhanuchhabra7634]

Sometimes I am amazed at how easy it is to follow when you explain, but my mind goes blank if I am left with some of the problems. At times I feel writing a program would be faster for me to solve such problems 😅. BTW, i attempted JEE in 2007.

[@paragggoyal1552]

i am well versed with probability, what the hell is the question saying??

[@WRSomsky]

So where do you get that "A is an event associated with a sample space S" means that A is a *set* of elements drawn from S? I have never heard this before. Without some prior indication of this interpretation, I would take it to mean that A is *an* element drawn from S.

[@sakumar]

In probability theory the definition of an event is "subset of the sample space." You could define an event as, say, "the number comes up even." It's the same as saying "the number comes up {2, 4, or 6}. So, event "odd" is {1, 3, 5}. Event "less than 5" is {1, 2, 3, 4}. Event "not 6" is {1, 2, 3, 4, 5} and event "prime" is {2, 3, 5} and so on. In this way, any subset of {1, 2, 3, 4, 5, 6} is an "event."

[@asdfasdfasdf1218]

That's just definition of "event" in probability. Probability theory uses words in ways that are very different from ordinary usage.

[@sleepheartcat]

"Sample space" is the set of possible "worlds" that might come to be. An "event" is something that in each world either happens or not. So you can define it as the set of worlds in which it does happen (i.e. a Boolean function on sample space).

[@anshumanagrawal346]

That's what an event from a sample space means.

[@Serg_144]

Another one "paradox" From the probability theory when you don't write precise enough the definitions... It depends on what the event is in this question There are 2 ways: 1) as in the video, event is {"we choose a subset of S"}. From that perspective the answer is correct. BUT there is another one which for me is more natural (Bernoulli scheme): 2) whe choose a subset from S, but SOME AMOUNT OF TIMES, not one, and also independently, e.g. we have A - subset, and B - subset, we have chosen TWO TIMES a subset from S, independently (!!) by definition. Now the event {A and B} from the 1) perspective is A intersect B. And from the 2) perspective is {"we choose A, and then we choose B"}. Here the answer is 1123 by the way, we simply don't get rid of some terms and summ them all, in comparison to the 1) P.S. kind of not nice to make the students solve the problem with 2 answers, when there is only one.... The problem isn't hard... Its just the fact that the authors didn't thought enough to make a normal task

[@ramunasstulga8264]

Jee aspirants never fail overcomplicate high school problems 🙊

[@akshatamritraj7399]

Jee is an exam for high school students though

[@duryodhan903]

I appeared in JEE advanced 2019, and was solving same paper yesterday to know how much can I score now after 5 years. But this question just exhausted me thanks presh to help me out😊😊😊.

[@Nepali_guy_Study_yt_account]

Did you get admission in IIT bro ???

[@N7492]

Cardinality sure looks like absolute value. An unclear question will result in unclear answers! It's the test writers who failed this one.

[@thekraken4265]

Currently a university student and my issue with the computer science, math and statistics courses I’ve taken is that the person making the questions doesn’t clarify things because I assume they think it’s trivial when it’s not.

[@prakhar77495]

This is not a probability question. Just a PnC question. Some questions are actually poorly worded in jee adv but well when they talk about no of elements it kinda becomes clear that they just want us to make subsets of S. This converts to sum of product of things taken two at a time from C(6,i) where i is from 1-6 you can find it using ((summation x)² - (sum of x²))/2. I guess many people would have messed up the calculations. Many people won't have seen the pattern and would have started calculating the expression without using above formula and then left it bcuz of the heavy calculations.

[@sundareshvenugopal6575]

Assuming two events are independent iff the outcome of one does not affect or influence the outcome of the other, that means if any two events share the same outcome it cannot be that they are unrelated and independent. The question now becomes, in how many ways can 6 elements of a set be split into two subsets such that one set has more elements than the other, taking into account symmetry of the two sets ?

[@thorstambaugh1520]

With a limited set and several elements impossible it is easier to bull through the combinations and total them

[@Aman-b5k]

This exam is written by the students of age 17-18

[@atharvsharma7648]

I don't get how are so many people criticizing the language of the question itself. I don't think anything was left unclear, indeed, such questions must be read a few times to digest all the details told in the question. There is nothing wrong with the way the question has been phrased, maybe there are better ways to phrase it, but it's telling all you need to know. An event is simply a subset of sets, thus the first thing that must pop into the mind would be that the only meaningful way to steer ahead would be to take an event as the subset of the provided set and move ahead.

[@nedmerrill5705]

I need to go to school on this video. I bet it took you longer than 3 minutes on this question...next question, please!

[@howareyou4400]

It's a normal question in probability class. But that's a undergraduate class...for physics and math major... So this is probably very hard for most high school students. Although I'm disappointed that there is no easier solution and we have to do the calculation with brutal force.

[@jpopelish]

A much easier question: There is a secret, 7 digit integer, made up of one copy of each of the digits, from 2 through 8. What must be added to the 4th digit, to make the secret integer divisible by 9.

[@Ibrahim208]

1

[@kilimanjarocruz660]

What does a probability of a subset even mean in this case? This seems to me to be extremely poorly written/defined.

[@anikchowdhury4959]

If |A|=2,3,4,5,6 and |B|=1,2,3,4,5 Then how |A intersection B|= 1,2,3,4,5 but not 2,3,4,5? 7:25

[@harikrishnareddygali6244]

I always have this question running in my mind when solving math problems like this " WHERE DO THESE QUESTIONS ORIGINATE?". Do people try to confuse others by asking a simple physical world question in mathematical terms? The language and nomenclature used in some math problems make them more complex to understand than they actually are.

[@timleader3253]

Have looked at the solution yet, but 10/36 seems like the obvious answer. B I >/= 1 is irrelevant, and of the 36 cases (think 2 dice) [B=1, A = 2, 3 ... 6, ][B = 2, A = 3, 4 ... 6][B= 3, A = 4, 5, 6][B= 4, A = 5,6][B=5, A=6]

[@timleader3253]

Well I was wrong

[@kiranvootori8101]

The only reason some could answer correctly is because they don't try to answer every question they get. Part of the test strategy is to find the questions they are strong in and then solve them. Not necessary that they will end up with the correct answer though.

[@sinohui3]

Don't forget the USMLE, a 9 hour exam.

[@varunsoni5278]

Whoever is struggling to understand the question should revisit how an event is defined mathematically

[@tan_k]

I think the majority of people who solved this question failed the test.

[@prabaln4752]

😂😂lol

[@pranshukumar7934]

Or Topped the test.

[@prabaln4752]

@@pranshukumar7934 that's not the case

[@armankumarsingh3566]

Yes😅

[@djjcyxz]

You know, it is simple. Basic probability theory. But the problem itself develops more roots as it progresses.

[@jasonryan2545]

If the cardinality of A has to be larger than that of B; along with the fact that cardinality is defined by the size of the set. How is the example of set A given here (2,3,4,5,6) larger than that of set B (1,2,3,4,5)? They both have the same number of unique elements in them, don't they?

[@coder6783]

The questions asked in JEE ADVANCED are not tough to solve. It’s just how you maintain your composure and stay calm while solving the problem. Once you decode the question and analyse calmly. It can be solved within few minutes.

[@Doeniz1]

It turned out to be not as hard as I expected when I heard how few people solved it.

[@Sgt_Bill_T_Co]

I went and made a cup of tea came back and moved on to something more useful.

[@dhruvalization]

Can you solve world hunger? \j

[@dolfindino7430]

78 kids did this is a mystery & paper leak is not new

[@Ibrahim208]

if there was even suspicion of paper leak in 2019, it would make national news for atleast 2-3 weeks

[@kiranvootori8101]

Just for information. Students appearing for JEE advanced are the top 10% of those who appear in JEE mains.

[@varunsoni5278]

That's not true. Why are you spreading misinformation?

[@rggrrggr290]

and important thing to mention is IT IS HIGH SCHOOL EXAM

[@rickyjuwono8000]

In my highschool i never learn cardinality

[@chair7728]

A lot of people are complaining about the wording being unclear, but the wording is completely unambiguous if you know the mathematical definitions, however I don’t know if the formal definitions are actually taught

[@N7492]

Vertical bars around a variable usually mean take the absolute value of the variable. Not its cardinality. The wording is indeed unclear.

[@chair7728]

@@N7492 the wording is not unclear as A,B are sets not numbers and it also states the exact meaning in the first statement

[@sandyalbanese8393]

I’m amazed that 78;people got this correct.

[@Indian_Ravioli]

i somehow could keep up with your explanation. but why didn't we use the same method as 4&3 or 3&2 for the first case of 6&1..that 62 made sense but cant it be achieved by the same method we got 180 in the other 2 cases?

[@potentialofprotonis3]

@9:47 when |A| = 2, shouldn't |B| = 3??? can u please explain why is it nothing

[@aakashshrestha6431]

|B|

[@ExcelInstructor]

9:38 How? i mean if A = 3 and we need result being multiple of 6, we have 3 * 2 = 6, but alse 3 * 4 = 12 which is another multiple of 6.

[@amanchandra2504]

B has to be less than A from the question which says that B is greater than 1 but less than A

[@SLuce222]

Someone, please help me understand! Like most people on here, I don't get their definition of an "independent event". I assumed the standard English language meaning that the 2 "events" are "independent". If A is an event in the sample space {1,2,3,4,5,6} (each equally likely). That would make 2^6 (64) possible outcomes {}, {1}, {2}, {1,2}, {3}, {1,3}...{1,2,3,4,5,6} and B is the same. So if there are 64 possibilities of A and B. The number of ordered pairs of (A, B) is 64^2 (4096), of which 1523 meet "1

[@chair7728]

I don’t know what you mean by the “English language meaning” but the independent they have used is extremely standard and widely used. It means that one event has no effect/does not tell you things about the other event. So for instance, two separate coin flips are independent since they don’t affect each other, but the events that you get a royal flush in poker and the event that you win the hand is not independent

[@SurprisedDivingBoard-vu9rz]

How many holes in the universe and how many burst and how many bubbles.

[@captwig]

‘You have 3 minutes to solve each question’. Then the video shows 15 minutes…

[@michaeldegrave5905]

The GRE being mentioned on the same list as some of these other tests is a laugh.

[@NaHBrO733]

The hardest part in this question is figuring out what is A and B. If you have a fundamental understanding of mathematical "events", you would know it is a subset of sample space. Then you have to know what is independent, and not confusing it with mutually exclusive. "A,B are subsets of S={1,2,3,4,5,6}, 1

[@NaHBrO733]

X is a random variable with sample space S. An event A is a subset of S. The probability of "A happening" is the probability of "the instance of X belongs in A"

[@bernardbruinius4951]

The only way to solve this problem in the allotted time, is you solved this question before.

[@ezxd5192]

What does "number of ordered pairs (A,B)" means

[@anshumanagrawal346]

I mean, if you're asking what an ordered pair is, it's just an element of P(S)×P(S) in this case (P(S) denotes the power set of S), satisfying the required condition. So you're essentially asked how many elements of P(S)×P(S) satisfy the given condition.