Let S be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an or

Q: Why is the total sample space $2^{10}$? I thought it was just $2^5$.

That is a common point of confusion! Remember, we are choosing an ordered pair $(A, B)$ from $P(S) \times P(S)$. Since there are $2^5 = 32$ possible subsets for $A$ and $32$ possible subsets for $B$, the total number of pairs is $32 \times 32 = 1024$, which is $2^{10}$. We are not just picking one subset; we are picking a pair of them.

Q: Why can't we just use combinations to solve this?

You absolutely can! You could sum the cases where $|A|=k$ and $|B|=m$ such that $k+m \le 5$. However, that leads to a tedious summation: $\sum_{k=0}^{5} \binom{5}{k} \sum_{m=0}^{5-k} \binom{5-k}{m}$. The element-wise approach is much more elegant because it bypasses the summation entirely by focusing on the independent choices of each element.

Q: Wait, why are there only 3 choices for each element? Shouldn't there be more?

Great question! Let's visualize the Venn diagram. An element can be in $A$ (but not $B$), in $B$ (but not $A$), or outside both. The only other possibility is being in both $A$ and $B$. But the problem explicitly states $A \cap B = \emptyset$, which means the intersection region is empty. Thus, that fourth choice is forbidden. We are left with exactly 3 valid choices.

Question

Let $S$ be a set of 5 elements and $P(S)$ denote the power set of $S$. Let $E$ be an event of choosing an ordered pair $(A, B)$ from the set $P(S) 	imes P(S)$ such that $A \cap B = \emptyset$. If the probability of the event $E$ is $\frac{3^p}{2^q}$, where $p, q \in \mathbb{N}$, then $p + q$ is equal to .........

Sigma Percentile · Accepted Answer

### The Elegance of the Element-Wise Perspective

Welcome, future engineer! Today, we are going to dismantle a beautiful problem in set theory and probability. It is the kind of problem that looks intimidating at first glance—a power set, an ordered pair, a disjoint condition—but once you peel back the layers, you will find a core of pure, crystalline logic.

### Phase 1: The Universe of Sets

Imagine you have a set $S$ containing five distinct elements: $S = \{x_1, x_2, x_3, x_4, x_5\}$. The power set $P(S)$ is the collection of all possible subsets of $S$.

For each element, we have a binary choice: it is either in a subset or it is not. With five elements, each having two choices, the total number of subsets is $2^5 = 32$.

We are picking an ordered pair $(A, B)$ from $P(S) 	imes P(S)$. Since there are 32 choices for $A$ and 32 choices for $B$, our total sample space size is:

$$32 	imes 32 = 2^5 	imes 2^5 = 2^{10}$$

### Phase 2: The Disjoint Constraint

The event $E$ requires that $A \cap B = \emptyset$. This is the heart of the problem. If you try to list all pairs $(A, B)$ that satisfy this, you will be counting until the next JEE exam!

Instead, let us shift our perspective. Let us stop looking at the sets $A$ and $B$ as monolithic entities and start looking at the individual elements.

Imagine each element $x_i$ as a traveler. In a standard Venn diagram, an element has four possible destinations: inside $A$ only, inside $B$ only, inside both $A$ and $B$, or outside both.

The condition $A \cap B = \emptyset$ acts like a security guard at the intersection region. It forbids any element from entering the 'both' zone. Each element now has exactly three valid destinations: it can join $A$, it can join $B$, or it can stay outside both.

### Phase 3: The Power of Independence

This is where the magic happens. Since the placement of each element is independent of the others, we can simply multiply the number of choices for each element.

We have five elements, and each has three choices. Therefore, the total number of favorable outcomes is:

$$3 	imes 3 	imes 3 	imes 3 	imes 3 = 3^5$$

This is the number of ways to form a pair $(A, B)$ such that they are disjoint. It is elegant, it is fast, and it is mathematically sound.

### Phase 4: The Final Comparison

Now, we bring it all together. The probability $P(E)$ is the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{3^5}{2^{10}}$$

The problem states that this probability is equal to $\frac{3^p}{2^q}$. By comparing our result, we immediately see that $p = 5$ and $q = 10$.

The final step is to calculate $p + q$. Adding these together, we get:

$$5 + 10 = 15$$

And there you have it! We have navigated the complexities of set theory and arrived at the answer with clarity and confidence. The final result is **15**.

Enter Numerical Value:

Visualized Solution

Defining the Set $S$

The Power Set $P (S)$

Total Sample Space

The Disjoint Condition

Element-wise Choices

Three Valid Choices per Element

Total Favorable Outcomes

Probability of Event $E$

Comparing with Given Expression

Final Answer: $p + q$

The Sigma Insight: Classical Definition of Probability

The Elegance of the Element-Wise Perspective

Phase 1

Phase 2

Phase 3

Phase 4

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