Animated Solution for Mathematics - Probability: If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

A Probability Odyssey

Imagine you are standing before a vast, shimmering vault containing exactly one hundred numbered spheres, labeled from $1$ to $100$. This is your sample space—your universe of possibilities.

In the realm of JEE Advanced, probability is not just about formulas; it is about understanding the structure of the universe you are operating within. Today, we are tasked with a specific mission: to reach into this vault and pull out three distinct spheres, such that every single one of them is divisible by both $2$ and $3$.

It sounds simple, but the beauty lies in the logic we use to navigate this selection.

Decoding the Condition

Many students stumble right at the start by trying to analyze divisibility by $2$ and $3$ separately. But let us pause and think like mathematicians.

If a number is divisible by $2$, it is even. If it is divisible by $3$, it is a multiple of $3$. To be divisible by both, a number must be a multiple of the Least Common Multiple of $2$ and $3$.

Since $2$ and $3$ are coprime, their LCM is simply $2\times 3=6$. This is our 'LCM Bridge'. We are no longer looking for two separate conditions; we have collapsed them into one elegant requirement: we are looking for multiples of $6$.

Our target set is now defined as $\{6,12,18,\dots ,96\}$.

Now, how many such numbers exist in our vault of $100$? We could list them, but that is tedious and prone to error.

Instead, we use the power of the floor function. We divide the upper limit of our range by our divisor: $n=\lfloor \frac{100}{6}\rfloor$.

Performing this division, we find $100÷6=16.66\dots$. Taking the floor, we get exactly $16$ favorable numbers. These are the only numbers in our universe that satisfy the condition. If we pick any three numbers from this subset of $16$, we are guaranteed success.

We have our total universe of $100$ numbers, and we have our favorable subset of $16$ numbers. We need to choose $3$ distinct numbers.

Because the order of selection does not matter—picking $6,12,18$ is the same as picking $18,6,12$—we must use combinations. The total number of ways to choose $3$ numbers from $100$ is ${}^{100}{C}_3$. The number of ways to choose $3$ numbers from our favorable $16$ is ${}^{16}{C}_3$.

Our probability $P(E)$ is the ratio of these two:

This is where many students panic, seeing the large factorials. But look closer. Recall that ${}^n{C}_r=\frac{n!}{r!(n-r)!}$.

When we write out the fraction, the $3!$ in the denominator of both the numerator and the denominator cancels out perfectly! We are left with:

Now, we simplify by grouping the terms: $\frac{16}{100}\times \frac{15}{99}\times \frac{14}{98}$.

1. $\frac{16}{100}$ simplifies to $\frac{4}{25}$. 2. $\frac{15}{99}$ simplifies to $\frac{5}{33}$. 3. $\frac{14}{98}$ simplifies to $\frac{1}{7}$.

Multiplying these together: $\frac{4}{25}\times \frac{5}{33}\times \frac{1}{7}$. We can simplify further by canceling the $5$ in the numerator with the $25$ in the denominator, leaving us with $\frac{4}{5\times 33\times 7}$.

Calculating the denominator: $5\times 33=165$, and $165\times 7=1155$. Thus, our final probability is $\frac{4}{1155}$.

We have arrived at the solution, but more importantly, we have mastered a process. Whenever you face a divisibility condition involving multiple numbers, always find the LCM first.

It simplifies the complexity of the problem instantly. Probability is not about guessing; it is about counting the right things in the right way. You have done exactly that. Keep this clarity of thought, and no JEE problem will ever be too daunting for you.