What exactly is the MECE principle?

Recently, I was thinking about Mutually Exclusive and Collectively Exhaustive events in probability theory. I tried to understand it through a simple problem and found some interesting observations that I would like to share. Let's get started!

Mutually Exclusive Collectively Exhaustive

In probability theory, We say a set of events are "Mutually Exclusive" if no two events can occur simultaneously. For example, when we toss a coin, the outcome can be either heads or tails but not both.

A set of events are said to be "Collectively Exhaustive" if at least one of these events will always occur. Let there be a set \( \,T = \{1,\, 2,\, 3,\, 4,\, 5,\, 6\} \), when we roll a die the outcome will always be one of the elements of \( \,T \). Therefore, we say the set \( \,T \,\) is collectively exhaustive.

When a set (of events) satisfy both of the above conditions we say it to be "Mutually Exclusive and Collectively Exhaustive". Let's write this definition more formally. Consider a set \(\,T = \{E_1, \,E_2,\, E_3\}\). The set \(\,T\) is said to be mutually exclusive and collectively exhaustive (wrt to a probability problem) if:

• \(\,E_1 \cup \,E_2 \cup \, E_3 = S \) (set of all possible outcomes) \( \rightarrow (1)\)
• \(\,E_1 \, \cap \, E_2 = E_2 \, \cap \, E_3 = E_3 \, \cap \, E_1 = \phi \rightarrow (2) \)

Why are we interested in this property? What is so special about it? If a set satisfies (1) and (2), then the following happens:

\( \begin{align} n(E_1 \cup E_2 \cup E_3) &=n(E_1) + n(E_2) + n(E_3) - n(E_1 \cap E_2) -n(E_2 \cap E_3) \\ & - n(E_3 \cap E_1) + n(E_1 \cap E_2 \cap E_3),\\ &=n(E_1) + n(E_2) + n(E_3) \quad using \, (2),\\ n(S) &= n(E_1) + n(E_2) + n(E_3) \quad using \, (1), \end{align} \)

\(n(E_1)\) = number of elements in the set \(E_1\).
If the set \(\,E_1 = \{3, \,7, \,8, \,10\}\), then \(n(E_1) = 4\).

Now you can see the initial equation is significantly simplified, which makes it easier to use in a probability or combinatorics problem. For example, if calculating \(n(E_1)\) is hard, you can simply calculate \(n(E_1) = n(S) - n(E_2) - n(E_3)\) which might turn out to be easier. The above derivation can easily be extended to a set having any number of events, making this result even more powerful.

Problem Statement

A box contains ten balls which are numbered from 1 to 10. Consider the experiment where we want to sample four balls with replacement, which means we would draw a ball, put the ball back into the box, and then draw the next ball. We repeat this process until we have drawn four balls.

Find the number of outcomes where:

1. Every ball has a different number
2. Exactly two balls have the same number
3. Exactly three balls have the same number
4. Every ball has the same number

Solution

To be more formal, let's define an event \(E_i\) which denotes the outcomes where exactly \(i \, \) balls have the same number. The problem statement basically asks us to compute \(n(E_1)\), \(n(E_2)\), \(n(E_3)\), and \ \, (n(E_4)\). Okay, let's do it!

1. We need every ball to be distinct, so during each draw, we shouldn't select the ball from the previous draw. Therefore, \(n(E_1) = 10 \times 9 \times 8 \times 7 = 5040\).
2. Out of the four draws any two draws should select the same ball which can happen in \({4 \choose 2} \times 10 \) ways. The remaining two draws should select a different ball which can happen in \(9 \times 9 \) ways. So, \( n(E_2) = {4 \choose 2} \times 10 \times 9 \times 9 = 4860 \).
3. Using the same approach as \( n(E_2) \), we get \( n(E_3) = {4 \choose 3} \times 10 \times 9 = 360 \).
4. We need every ball to be the same, so during each draw, we should select the ball from the previous draw. Therefore, \( n(E_4) = 10 \times 1 \times 1 \times 1 = 10 \).

Did we do it? Is the above approach correct? The explanations seem reasonable, right? Let's try to test our solution against some well-known conditions.

Verifying the Solution

A simple test we can do is to check if each of our calculated outcomes is less than the total number of outcomes. That is, if \(n(E_i) < n(S) \; \forall \; i \), where \(S \) is the universal set. Calculating \( n(S) \) is easy. Each of the four draws has \(10\) possible outcomes, therefore \(n(S) = 10 \times 10 \times 10 \times 10 = 10^4 \).

Since each \( n(E_i) \) is less than \( 10^4 \), our solution passes this test. Can you guess what our next test will be? You are correct if you guessed Mutually Exclusive and Collectively Exhaustive. Else why would I explain it earlier :)

The way the events \( E_1 \), \( E_2 \), \( E_3 \), and \( E_4 \) are defined makes them MECE. Hence the following test needs to pass.

\[ n(E_1) + n(E_2) + n(E_3) + n(E_4) = n(S) \]

Remember, we derived this equation in the MECE section of this blog.

The required sum is \( 5040 + 4860 + 360 + 10 = 10,270 \) which is clearly greater than \( n(S) \). So we might have miscalculated one or more \( n(E_i) \)'s.

Where is the mistake?

Before I explain the mistake, I recommend you go back to the solution and try to find it. Since our required sum is greater than \(n(S)\), we might be double counting a few permutations.

The issue lies in \(n(E_2)\)'s computation. Our approach double counts the cases when two pairs of draws choose the same number. For example, 1122, 1313, 4466, and similar draws will be double-counted. Let's try to understand why this happens.

1 1 2 2 means \(1^{st}\) draw \( \rightarrow \) 1, \( \, 2^{nd}\) draw \( \rightarrow \) 1, \( \, 3^{rd}\) draw \( \rightarrow \) 2, and \( \, 4^{th}\) draw \( \rightarrow \) 2.

Assume we are just starting the experiment, so we haven't chosen any balls yet. We will be choosing \( {4 \choose 2} \times 10 \times 9 \times 9 \) times such that exactly two draws have the same ball (or number). This process will double count 1122.

1. Frist Permutation

   • start the draw \( \qquad \qquad \qquad \qquad \quad \rightarrow \, \, \) ― ― ― ―
   • two draws picks same ball, \({4 \choose 2} \times 10 \enspace \rightarrow \, \, \) 1 1 ― ―
   • remaining two draws, \(9 \times 9 \qquad \quad \enspace \, \, \rightarrow \, \, \) ― ― 2 2
   • one permutation complete \( \qquad \qquad \enspace \rightarrow \, \, \) 1 1 2 2

2. Second Permutation

   • start the draw \( \qquad \qquad \qquad \qquad \quad \rightarrow \, \, \) ― ― ― ―
   • two draws picks same ball, \({4 \choose 2} \times 10 \enspace \rightarrow \, \, \) ― ― 2 2
   • remaining two draws, \(9 \times 9 \qquad \quad \enspace \, \, \rightarrow \, \, \) 1 1 ― ―
   • one permutation complete \( \qquad \qquad \enspace \rightarrow \, \, \) 1 1 2 2

Now, I hope you understand why 1122, 1313, 4466, and similar permutations would be double-counted. So how do we correct it?

Corrected Solution

We can correct this issue by aggregating the double-counted permutations and dividing them by 2. So our new way of calculating \( n(E_2) \) becomes:

\( \begin{align} n(E_2) &= n(only \; one \; pair \;o\!f \;draws \; has \;same \; ball)\\ &+n(two \;pairs \;o\!f \;draws \; has \;same \; ball), \\ &= {4 \choose 2} \times 10 \times 9 \times 8 + \frac{{4 \choose 2} \times 10 \times 9 \times 1}{2},\\ &= 4320 + 270,\\ &= 4590, \end{align} \)

There are \( {4 \choose 2} \times 10 \times 9 \times 1 \) double-counted permutations

Let's try to apply the MECE test on our new solution to check its correctness. Sum of \(n(E_i)\)'s = 5040 + 4590 + 360 + 10 = 10,000 = \(n(S)\). Our new solution passed the test! We have finally solved the problem.

Conclusion

I hope you learned something valuable from this blog. Suppose you are still confused or would like to try a similar problem. I suggest you solve the same problem with six draws instead of four, making things more interesting. Suppose you have any questions feel free to comment below or DM me. I would be happy to help.

Before I finish this article, I will leave you with a question to ponder. In the verification section, I said that \(E_1\), \(E_2\), \(E_3\), and \(E_4\) are MECE due to their definition. Can you see them satisfying the two required properties?

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