Animated Solution for Mathematics - Probability: Let $S=\{1,2,\dots ,20\}$. A subset $B$ of $S$ is said to be "nice", if the sum of the elements of $B$ is 203. Then the probability that a randomly chosen subset of $S$ is "nice" is:

My dear student, welcome to a problem that, at first glance, seems designed to break your spirit. You are presented with a set $S=\{1,2,\dots ,20\}$ and asked to find the probability of choosing a 'nice' subset—one where the elements sum to exactly $203$.

If you start by trying to list combinations that add up to $203$, you will be there until the next century. But here is the secret of the JEE Advanced: the most difficult-looking problems often have the most elegant, simple exits. We just need to find the right door.

The Total Sample Space

Before we dive into the 'nice' condition, let us establish our universe. We have a set $S$ containing $20$ distinct elements.

For every element in this set, we face a binary choice: do we include it in our subset, or do we leave it out? Since there are $20$ elements and each choice is independent, the total number of possible subsets is simply $2\times 2\times \cdots \times 2$ ($20$ times).

Mathematically, our sample space size is $2^{20}$. This is our denominator. Now, we just need to count the 'nice' subsets.

The Complementary Insight

Here is where the magic happens. We are looking for a subset $B$ such that the sum of its elements is $203$. But look at the total sum of all elements in $S$.

Using the arithmetic series formula, $S_{total}=\frac{n(n+1)}{2}$, we calculate:

Think about this. If the total sum of all numbers is $210$, and our 'nice' subset $B$ sums to $203$, what does that imply about the elements we excluded? Let $B^c$ be the complement of $B$ (the set of elements not in $B$).

The sum of elements in $B^c$ must be:

Do you see the beauty of this? Instead of hunting for a massive sum of $203$, we are hunting for a tiny sum of $7$. Every time we find a subset that sums to $7$, we have automatically found a 'nice' subset that sums to $203$. It is a perfect, one-to-one mapping.

The Systematic Hunt for 7

Now, we must be disciplined. We need to find all subsets of $S$ whose elements sum to $7$. We will categorize them by the number of elements in the subset.

Case 1: Subsets with 1 element This is trivial. We need a single number that equals $7$. The only candidate is $\{7\}$. That is $1$ subset.

Case 2: Subsets with 2 elements We need two distinct numbers $x,y\in S$ such that $x+y=7$. Let us list them systematically to ensure we miss nothing:

- If $x=1$, then $y=6$. (Valid: $\{1,6\}$) - If $x=2$, then $y=5$. (Valid: $\{2,5\}$) - If $x=3$, then $y=4$. (Valid: $\{3,4\}$)

If we try $x=4$, we get $y=3$, which is the same set. So, we have $3$ subsets here.

Case 3: Subsets with 3 elements We need $x+y+z=7$. Let us pick the smallest possible values to keep the sum low.

If we take $1$ and $2$, their sum is $3$. To reach $7$, we need $4$. So, $\{1,2,4\}$ works.

Are there any others? If we try $\{1,3,x\}$, then $1+3=4$, so $x=3$. But elements must be distinct! We cannot use $3$ again. Thus, $\{1,2,4\}$ is the only $3$-element subset. That is $1$ subset.

Case 4: Subsets with 4 or more elements Let us test the smallest possible sum for $4$ elements: $1+2+3+4=10$. Since $10>7$, it is impossible to form a sum of $7$ with $4$ or more elements. Our search is complete.

The Final Calculation

We have found our favorable outcomes: - $1$ subset of size $1$ - $3$ subsets of size $2$ - $1$ subset of size $3$

Total favorable outcomes = $1+3+1=5$.

Finally, the probability $P$ is the ratio of favorable outcomes to the total sample space:

There it is. We navigated the complexity by changing our perspective, turning a daunting sum of $203$ into a manageable sum of $7$. This, my friend, is the essence of mathematical maturity. You didn't just solve a problem; you mastered the logic behind it. Keep this mindset, and no JEE problem will ever be able to stand in your way.