# Probable Fallacies

After the recent paper of JEE-MAIN 2015, a lot of people have been arguing with me about a question that was present in the paper. Before we look into the problem, first, let me give you another simple problem:

1.1 Two different coins are tossed simultaneously, find the probability that both show heads.

1.2 Two identical coins are tossed simultaneously, find the probability that both show heads.

Obviously, the answer to 1.1 is $1/4$ which is evident as we have one favorable case in the sample set $\{ (HH),(HT),(TH),(TT)\}$.

Continuing on similar lines, people argue that the answer to 1.2 is $1/3$ as the sample set contains only : $\{ (\text{two heads}),(\text{one head \& one tail}),(\text{two tails})\}$ and we have one favorable case.(but this reasoning is incorrect. Infact the answer is still 0.25)

We now come to the actual problem:

2.If $12$ identical balls are to be placed in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls is I'm not giving the answer or the method to do this problem as I want the reader to try it on their own before joining the discussion. The reader is advised to think about given problems before looking at the comments below.

Please do help me solve this! Thanks! Note by Raghav Vaidyanathan
5 years, 6 months ago

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I think that the correct answer to this problem is not given in options which makes it a wrong question in JEE-Mains.

- 5 years, 6 months ago

This problem looks ambiguous. "the probability that one of the boxes contains exactly $3$ balls is" implies that exactly one box have exactly $3$ balls? Or at least one box have exactly $3$ balls?

- 5 years, 6 months ago

I think it means: "probability that there are exactly three balls in exactly one of the boxes".

- 5 years, 6 months ago

No, what I'm trying to say is that "Can there be another $3$ balls in another box?" Do we take into account for $(3,3,6)$?

- 5 years, 6 months ago

No, we don't consider $(3,3,6)$. But I'm not exactly sure if that's the case. Try both ways, if possible! Thanks

- 5 years, 6 months ago

Either way, I'm not getting the answer in any of the choices given.

I assume the probability of assigning the balls is unbiased and we can have no balls in one or two boxes. And it doesn't matter which box actually have $3$ balls in them.

The number of possible scenarios is equivalent to the number of ordered triplets of non-negative integers for the diophantine equation: $a_1 + a_2 + a_3 = 12$, a simple use of stars and bars gives $\dbinom {12 + 3 - 1}{3 - 1} = 91$.

And the unordered triplets of solutions such that exactly one of $a_1,a_2,a_3$ is equals to $3$ is $(3,0,9), (3,1,8), (3,2,7),(3,4,5)$, thus its ordered pairs is simply $4 \times 3! = 24$.

Take their ratio yields $\frac {24}{91}$. If we account for the fact that $(3,3,6)$ is another unordered solution, then there's $\frac {3!}{2!} = 3$ more solutions, thus giving $\frac {24+3}{91} = \frac {27}{91}$.

- 5 years, 6 months ago

Yeah, I did it in a similar way during the exam. I was not able to answer the problem. I think the confusion arises due to the fact that they have not mentioned the mechanism by which the random distribution is done. If it is done by assuming that each ball can go to each box with equal probability, then your answer(the one above) will be incorrect as some of the cases that you have considered will be more probable than the others.

- 5 years, 6 months ago

I have a feeling the question does assume that each ball can go in each box with equal probability, instead of assuming that each configuration is equally possible. But even if you assume that, you still won't get the answer. I've tried that and although my answer is in a similar form, it isn't actually the same.

- 5 years, 6 months ago

I believe the question assumes the fact that probability of happening of an event does not depend on the fact of our ability to distinguish between the objects.

I mean to say that the probability of the given event should not change even if we mark every box and ball so that each one is distinguishable then also the probability of the event should not change.

- 5 years, 6 months ago

- 5 years, 6 months ago

That question from JEE MAIN 2015 is incorrect. The answer to question 1.1 and 1.2 is $\dfrac{1}{4}$ only. The given cases in 1.2 are not equally favorable. Also, I don't see any link between the questions above and JEE MAIN 2015 Problem.

- 5 years, 6 months ago

Does the probability under consideration change if we assume the balls and boxes to be distinct even though they are given to be identical?

It doesn't change in the coin case, hence i mentioned it so that people don't disagree with me outright without listening to my argument.

- 5 years, 6 months ago

The two cases of box/balls and coin are totally different. And yes, the probability changes. See how:

Case I: Consider you have $2$ identical boxes and $2$ identical balls. Possible cases would be $(2,0),(1,1)$

Case II: Consider you have $2$ distinct boxes and $2$ identical balls. Possible cases would be $(2,0),(1,1),(0,2)$

Case III: Consider you have $2$ identical boxes and $2$ distinct balls. Possible cases would be $(2,0),(1,1)$

Case IV: Consider you have $2$ distinct boxes and $2$ distinct balls. Possible cases would be $(2,0),(1,1),(1,1),(0,2)$. To clarify, call the balls $A$ and $B$, cases would be $(AB,0),(A,B),(B,A),(0,AB)$

- 5 years, 6 months ago

I agree with you on the possible cases. But how can you claim that they are equiprobable? Can we just define the possible cases for each Case to have equal probability? Are you saying that while distributing the balls among the boxes, we do it in such a way that the said cases are equally probable?

- 5 years, 6 months ago

Yeah, that's great demonstration. @Pranjal Jain

- 5 years, 6 months ago

Thanks! :)

- 5 years, 6 months ago

I'll prove the probability of getting 2 balls in a box is the same as getting 1 ball in each, irrespective of whether the balls/boxes are distinct.

We take 1 ball and put in any box. Now, we see that the second ball can go into two boxes. The box with the ball, or the box without one. If I put it in the former, I get a box with 2 balls, and if I put in it the 2nd, I get two boxes with one ball each. Since each case is equiprobable, that implies getting a box with 2 balls is equivalent getting each 2 boxes with a ball each.

The reason why your case II is wrong, is that $(1,1)$ and $(0,2)$ occur with unequal probabilities.

It is better to be rigorous in a situation in which there appears to be a fallacy. What is the question in your cases? The best bet would be "Each ball can be put into a box with equal probability. What is the probability one box contains 2 balls? " If you interchange ball with coin and putting in a box with flipping a coin, we get the exact same case as the coins case Raghav gave.

- 5 years, 6 months ago

Yes! Finally! The method of distribution is under the condition that each ball can be put into any one box with equal probability! Thus making this same as the distinct case!

- 5 years, 6 months ago

I never said that they are equiprobable. I just started listing out cases.

- 5 years, 6 months ago

Then what was the purpose of listing out the cases?

- 5 years, 6 months ago

Off course, the probability will change if you assume the balls and boxes to be distinct even though they are given to be identical. You can check this by assuming 2 balls and 2 boxes so that you can count all the possible cases manually. And hence find your answers by taking both the situations one by one.

- 5 years, 6 months ago

Consider the case of throwing a single coin one at a time

In the case of coins, each throw is different ,you can classify as 1st toss, 2nd toss, 3rd toss and so on,,

so after 5 trials, the case of

(H,T,T,T,T) is different from (T,H,T,T,T)

However, if the throws were not distinct , Or you cannot tell the difference between throws , your sense of time is lost

then

(H,T,T,T,T) is same as the other

now consider the throws as boxes, and you will see that the indistinguishability of boxes does matter

Now, some one might say that the ratio must remain constant as both the sample space and the number out comes decrease, true that both decrease but they do not decrease proportionally,

for example, you can get two heads and 3 tails in 5C3 ways and upon announcing the nature of identicality, they all represent just 1 outcome,.

but you can get 5 heads in just 1 way,

(basically, different macrostates have different number of microstates, hence you cannot claim that the sample space and the number of outcomes decrease proportionally to maintain the probability)

- 5 years, 6 months ago

You are right in claiming that they don't decrease proportionally. What I think happens is that the probability of said events increases proportionally. That is, the events described in the sample space that you mentioned are not equiprobable , hence we cannot use our "count favorable and divide by total" method.

Let's do your analysis for three coin tosses. We will compare the values we obtain when we assume identical and when we don't. Let the event identical be denoted as $Id$ and distinct be denoted as $Di$

In $Di$, there are $8$ equiprobable events(each with probability $1/8$). It's easy to list them out.

Now let us forget about time and assume $Id$. In the sample space of $Id$, there will be lesser number of events than in $Di$. This is easy to see, as events like $(H,T,T)$, $(T,T,H)$, $(T,H,T)$ are identical now. In $Id$ These three events will represent one event which is: $(\text{one head and two tails})$. But the thing is that:

$P((\text{one head and two tails})|Id)=P((H,T,T)|Di)+P((T,T,H)|Di)+P((T,H,T)|Di)=3/8$

Similarly $P((\text{three heads}))=P((H,H,H)|Di)= 1/8$

Hence, we find that the number of elements in the sample space have decreased in a non proportional way, but the probability has increased to account for this.

I'm convinced that this is right. @Mvs Saketh . This seems to be true in the coins case, but as @Pranjal Jain has said below, distribution of balls into boxes seems to be slightly different.

- 5 years, 6 months ago

are you saying that the questionis right?

Bro, If you randomnly throw balls into boxes, surely, upon throwing a thousand times, you will get the same distribution whether or not the boxes are different or not (as physical events have nothing to do with nature of boxes, nature can differentiate them even if we cant)

But, as far as mathematics is concerned, we are to stick to the classical definition of probability where probability means, fraction of the favourability of the event, and hence there is nothing wrong with using probability equation.

Infact we do it all the time, like how many ways can we select out of

8 apples , 4 bananas and 2 oranges so that atleast 1 orange is present,

and if someone asks what is the probability of such a choice

you will say 2/3 (which you can easily check)

however, if i say that the fruits of same type are different, then your answer will change

Now imagine that i tell you to make the selections and secretly mark each orange different from each other (say by putting different radioactive substances in each) and so on, and yet as far as you are concerned the probability is 2/3 though for me it is different since i know they are distinct

So, based entirely on the given situation, we can safely claim that since each box is identical to me, and hence i can assume each macrostate to have just one microstate for me and are equiprobable and claim that the options are incorrect

(This leads to a reasonable paradox if you try to make things more physical, does the same experiment give different results based on the nature of objects? Yes , if a human is told to do the arrangement who cannot differentiate between them but NO if nature acts upon them, because it can differentiate among them no matter how identical they are)

- 5 years, 6 months ago

Let us perform the above mentioned fruits experiment on a smaller scale: with $3$ apples and $2$ oranges.

Before we continue, I would like to make sure if I have understood you correctly:

1. If I say that all fruits of a type are identical, the probability that a selection of fruits contains at least one orange is $2/3$.

2. I think that all the fruits of one type are identical, but without my knowledge, you have labelled the oranges with radioactive tracers such that only you can differentiate between the two oranges. In this case, the probability that a selection of fruits contains at least one orange will remain $2/3$ for me, but it will be $3/4$ for you?

Reply to this comment, and I will post my final comment which will remove the confusion once and for all.

- 5 years, 6 months ago

yes but note that it is not as per the physical definition of likelihood but by the mathematical definition of probability, and yes , i think i know what you will say :) , that each macrostate is not equally probable, but i think that depends on what basis you select the fruits. please proceed

- 5 years, 6 months ago

I guess you already know what I'm going to say :). But yeah, here it is:

It will suffice to try to convince you that the probability for you is not $3/4$ but is actually $2/3$. And yes, It depends on the basis of selection of fruits. The first bag of fruits(apples) is immaterial to the question. So we will concentrate on the oranges only. You can verify using your arguments that the probabilities remain $2/3$ and $3/4$ even now.

While choosing from the bag of oranges, for me, the oranges look like: $(O,O)$. But for you they look like $(o,O)$. I am tasked with making a selection of oranges and the question is to find the probability that I will choose at least one orange.

What I observe: I can select no orange, one orange, or two oranges. The favorable case is when I choose one orange or two oranges. Now how do I find the probability? I can simply divide the number of favorable cases by total number of cases, since for me taking none, one or two is all the same. Hence my answer would be $2/3$

What you observe: I can select no orange, select $o$, select $O$, or select both. You think that I'm equally inclined to perform any one of the above four actions, but in reality I am not. If you write down the selection that I make each time, and then look at it after, say $3000$ experiments. You will observe that I picked no orange about $1000$ times, I picked both oranges $1000$ times and I picked only $o$ $500$ times and only $O$ $500$ times. Hence, the probability that I pick at least one orange is still $2/3$.

So you see here(as you already know) that I'm claiming that your sample space in the radioactive experiment is not made of equiprobable elements. I am twice as likely to select both oranges when compared to selecting just $O$. I am also twice as likely to select no orange when compared to selecting just $o$.

- 5 years, 6 months ago

What about that differential equation question??

- 5 years, 6 months ago

it is correct, assuming dy/dx exists at x=1 (which is a reasonable assumption and can be proved) , put x=1 in the given differential equation, you will get y=0 which means the curve passes through (x,y)=(1,0) with this we get the integration constant as c=2,

- 5 years, 6 months ago

Can u pl. Show the proof of why only x=1...

- 5 years, 6 months ago

it is already given that x>=1 , you can check that it cannot hold true for any x>=1 , there are only two values 0 and 1 for which y=0

- 5 years, 6 months ago

why do we want y to be 0

- 5 years, 6 months ago

but i guess it comes out to be like y=(constant) now the constant can be any value

- 5 years, 6 months ago

The equation we get is:

$y log(x)= 2x(log(x)-1)+C$

Putting $x=1$ gives:

$0=-2+C \Rightarrow C=2$

- 5 years, 6 months ago

ok after reading the above solution please answer 1) what was the logic of putting x=1 here , why not anything else ?

yln x=2(x*ln x-x ) +c, y=2(e-e)+c, y=c

- 5 years, 6 months ago

Usually when we solve differential equations, we get a general form in $x$ and $y$ which involves one or more constant terms which are arbitrary. Hence we get different solutions for different values of these constant terms. Usually, the value of $y$ for some value of $x$ will be given, so that we can find the constants in the solutions without any ambiguity. Take this as an example:

1. Solve $\frac {dy} {dx}=2x$. Given that the curve passes through $(0,10)$.

The solution is obviously $y=x^2+C$, but since it passes through given point, $C$ can only take one value, which is: $C=10$.

Similarly, in our problem, we finally get: $y log(x)= 2x(log(x)-1)+C$. But no other information about curve(like some point that it passes through) is given. Therefore, at first look, it seems that there is no condition on $C$ and hence we have many solutions, since it can take any value. But this is not so. The equation must be consistent for all $x$. Therefore, even at $x=1$, L.H.S must be equal to R.H.S. That is, if $C \ne 2$, the equation becomes false at $x=1$, which is a contradiction. We realize that there does exist a condition on $C$, but it is hidden in a subtle manner. The idea for putting $x=1$ inspired by the fact that the $y$ term disappears when $x=1$.

- 5 years, 6 months ago

So I'm guessing your solution goes something like this:-

Let the balls and boxes(A,B,C) be distinct.

Then number of ways Box A can can have 3 balls is ${12 \choose 3} * 2^9$. We choose 3 balls to put in A and put the remaining 9 balls in the two boxes in $2^9$ ways.

The same can be done for box B and C. Next we remove over counted cases, of which there are three. So total number of cases are $3* {12 \choose 3 }* 2^9 - 3$. Total number of cases is $3^{12}$. Therefore probability is $\dfrac{3* {12 \choose 3 }* 2^9 - 3}{3^{12}} =\dfrac {3*220*2^9-3}{3^{12}} = \dfrac{3*55*2^{11} - 3}{3^{12}}$

For the sake of trying to get an answer, I remove the 3, or basically add back the double counted cases,

$\dfrac{3*55*2^{11} }{3^{12}} = \dfrac{55*2^{11} }{3^{11}} = 55*\left (\dfrac{2}{3}\right )^{11}$

This now makes me look like an idiot since it isn't an answer.

- 5 years, 6 months ago

@Raghav Vaidyanathan Shouldn't the answer to problem 1.2 be $\frac {1} {4}$? If we first throw coin 1 and then coin 2 (this should be the same as throwing coins simultaneously, because each outcome is decided individually) the fact that they are identical won't change the probability!

- 5 years, 5 months ago

I am sorry Andrei, it seems like someone has edited this note. I had clearly mentioned that the answer to question 1.2 was same as that of 1.1. I have clarified this fact once again. Thanks for your comment!

- 5 years, 5 months ago

First let me tell you that the word "identical" has no such significance if we speak about real world objects in combinatorics (unlike letters, numbers, etc.). Thus the balls are though seemingly identical, they are physically distinct. However, if, in this problem, we consider the boxes to be physically distinct, then the answer that comes up doesn't match with any of the provided ones. So, instead of the boxes, we consider three indistinguishable theoretical "holes". Then, in one case, if the contents of such holes are given by (3,5,4), and in another case (3,4,5), then the two cases are indistinguishable. Now suppose we are blindfoldedly throwing the balls one by one, such that each ball will land up inside a unique box, and no ball misses all the three boxes. Then each ball has three options, and the twelve throws being independent, there are altogether 3^12 equiprobable ways. Now in the previous discussions, the question has been raised that whether "one box" means "exactly one box". or "at least one box". It has not been explicitly mentioned though, yet it is expected for one to assume that the latter condition has been talked of in the problem. This is because in the very same sentence, they mention "exactly three balls". So, from the 12 balls, these 3 balls can be chosen in $\left( \overset { 12 }{ \underset { 3 }{ } } \right)$ ways. For each such way, (12-3)=9 balls and (3-1)=2 boxes remain, and these boxes can be filled up in 2^9 ways. So, let the event be E. So, using the conventional notations of probability: $n(S)={ 3 }^{ 12 }$ $n(E)=\left( \overset { 12 }{ \underset { 3 }{ } } \right) .{ 2 }^{ 9 }$ $P(E)=\frac { n(E) }{ n(S) } =\frac { \left( \overset { 12 }{ \underset { 3 }{ } } \right) .{ 2 }^{ 9 } }{ { 3 }^{ 12 } } =\frac { 220.{ 2 }^{ 9 } }{ 3.{ 3 }^{ 11 } } =\frac { 55.{ 2 }^{ 11 } }{ 3.{ 3 }^{ 11 } } =\frac { 55 }{ 3 } .{ \left( \frac { 2 }{ 3 } \right) }^{ 11 }$

- 5 years, 3 months ago