Made Easy Probability - Let X be a set containing n elements. Three subsets A,B, C of X are chosen at random. The probability that A, B, C are pairwise disjoint is? (What do they mean by pairwise disjoint? and how should I approach this question?)

Question

Comments

From ChatGPT – which solution is better? :)

"Pairwise disjoint" means that no two sets have a common element, i.e., A ∩ B = ∅, B ∩ C = ∅, and A ∩ C = ∅.

The problem is a bit tricky since the subsets are chosen randomly. In general, the probability can be quite complex to calculate and may involve advanced combinatorial techniques.

But if we simplify things and suppose that each element of X independently decides which of the sets A, B, and C to go into, then the answer is relatively simple:

1) Each element has four choices: to go into none of the sets, or to go into one of A, B, or C. So there are 4^n ways to distribute n elements into {A, B, C, ∅}.

2) If A, B, and C are pairwise disjoint, then each element has two choices: either go into ∅ or go into one of the sets {A, B, C}. So there are 2^n ways to distribute n elements into disjoint {A, B, C, ∅}.

Therefore, the probability that A, B, and C are pairwise disjoint is (2^n) / (4^n) = (1/2)^n.

Please note: This is a simplification that may not apply if the subsets A, B, and C are not chosen independently or if other constraints are in place. Sometimes, the only feasible way to find the solution is to use more sophisticated combinatorial methods.

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I did not understand what the second point was saying – 

If A, B, and C are pairwise disjoint, then each element has two choices: either go into ∅ or go into one of the sets {A, B, C}. So there are 2^n ways to distribute n elements into disjoint {A, B, C, ∅}.

 The way I interpreted the given solution was – 

1. 1. We have 4 possibilities for each element (to go into none of the sets, or to go into one of A, B, or C.) So there are 4^n ways to distribute n elements into {A, B, C, ∅}. This becomes the favourable number of events from our sample space

 

2. Now coming to the sample space, we have 2^n choices to form set A, 2^n choices to form set B and 2^n choices for set C. This makes our sample space contain (2^n)^3 elements = 8^n elements.

Now, applying the basic formula of probability – favourable/Total, we got the answer as – 4^n/8^n = 1/2^n

I wonder if I am understanding it right. @arjun, can you please elaborate on ChatGPTs solution please.

Answer 1

lets go to the basics:
we have $n$ elements, what are the total ways we can distribute them in A,B,C (no restrictions).

for each element x in X,

1. x not in any of A,B,C                  –  1 way
2. x is exactly in one of A,B,C        –  3 ways
3. x is in any two of A,B,C              –  3 ways
4. x is in all A,B,C                           – 1 way

Total ways we can distribute each element is 8, number of ways we can construct A,B,C is $8^n$.

 

only in case of 1. and 2., A,B,C will be disjoint, hence we have 4 ways for each element
total ways in which A,B,C are disjoint is $4^n$

required probability = $\frac{4^n}{8^n} = \frac{2^{2n}}{2^{3n}} = \frac{1}{2^{n}}$

Comments

I agree with you, we should not assume replacement here by default.
As you mentioned Question is bit tricky to solve in case of without replacement, will think of a better way to solve this.

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now given it some more thoughts, It is not so straight forward to solve, even difficult to solve as n grows, and likely to not find any closed form.

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Yes, exactly. Unfortunately not many GATE aspirants understand this :(

Answer 2

I did not understand what the second point was saying – 

If A, B, and C are pairwise disjoint, then each element has two choices: either go into ∅ or go into one of the sets {A, B, C}. So there are 2^n ways to distribute n elements into disjoint {A, B, C, ∅}.

 The way I interpreted the given solution was – 

1. 1. We have 4 possibilities for each element (to go into none of the sets, or to go into one of A, B, or C.) So there are 4^n ways to distribute n elements into {A, B, C, ∅}. This becomes the favourable number of events from our sample space

 

2. Now coming to the sample space, we have 2^n choices to form set A, 2^n choices to form set B and 2^n choices for set C. This makes our sample space contain (2^n)^3 elements = 8^n elements.

Now, applying the basic formula of probability – favourable/Total, we got the answer as – 4^n/8^n = 1/2^n

I wonder if I am understanding it right. @arjun, can you please elaborate on ChatGPTs solution please.

Comments

Two questions

1. Just considering the denominator or the total cases - you considered sets are being selected by replacement right? Otherwise we have 2^nC3 possibilities.

2. For the favorable cases, for any element it has 4 possibilities as mentioned. But how can we say that these 4 possibilities are equally likely? Otherwise their probabilities won't be the same.

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Yes, I assumed that sets are selected by replacement as all three A, B, and C are just subsets of the given set X, so individually they have 2^n possibilities. And since nothing is mentioned in the question I took the general number of ways to make a subset.

Coming to the second point, again since nothing is mentioned, we are considering that all 4 cases are equally likely, otherwise, it should have been explicitly mentioned. (Similar to the case when we solve questions based on a biased coin/dice)

My doubt here is, how are we getting the probability as 2^n/4^n in ChatGPT’s solution? Though the answer remains the same, which approach is correct?

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If I ask you “In how many ways you can select 3 subsets for an $n$ element set” what will be your answer? – Replacement should not be the default here.

The best part of the ChatGPT answer is this line

But if we simplify things and suppose that each element of X independently decides which of the sets A, B, and C to go into, then the answer is relatively simple:

Basically, it assumed one thing and the given solution assumed another thing. In other words both the solutions are for different questions and not for the one originally asked. The original question is not easy to solve (no generic solution) and you can ask this in stackexchange for more details if you’re interested.

PS: The point of practice questions is to ensure you do well in actual exam questions. But by learning such solutions aspirants are only increasing their chance of negative marks in GATE. 

Just another reply from ChatGPT to the same question:

I'm sorry, but as an AI language model, I currently lack clear mathematical notation capabilities to describe the precise solution to this problem in its entirety.

However, I can help guide you on how to approach it:

Being pairwise disjoint means that no two of the three sets share any common elements. With replacement means that after choosing a subset, all of its elements are returned to the set X before the next subset is chosen.

Because of the complexity of the problem and elements at play, it may be difficult to generate a straightforward formula for the probability without making certain assumptions.

Consequently, it might be best to approach this problem using programming or simulation for a precise numeric answer. Alternatively, you could consider simple cases first (like n=3 or 4), look for patterns, and then try to generalize.

Generally speaking, the probability will be heavily dependent upon both n and the sizes of the subsets A, B, C. If you have more information about either of these, it would significantly simplify the problem.

 

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Thank you for such a clear explanation, sir.

If you had not emphasized enough on the right approach towards such questions, I would have wasted more time on it.

Answer 3

For favourable :

Each element has 4 ways , So 4^N

Total ways :

For each subset , it can either have 0 elements or 1 element or 2 elements .. or n elements which can be done by

1 + NC1 + NC2 … NCN ways = 2^N ways

For 3 subsets A,B and C Total ways = (2^N)^3= 8^N

Favourable outcomes = 4^N/8^N

Answer 4

We can also solve  this by small example

let take n=2

X={1,2}

hence all possible subset ={phi,{1},{2},{1,2}}

so all  favourable case

                                     A      B        C

                                     phi   {1}          {2}=>6 cases

                                     phi    phi         {1}=>3  cases

                                    similary for {2}

                                    phi    phi         {2}=>3 cases

                                    phi    phi          {1,2}=>3 cases

                                    phi,  phi            phi =>1 cases

 so total  =16 and  all possible cases 4*4*4

so probability= 16/64    =¼  hence option B is correct which is 1/2^n