Animated Solution for Mathematics - Probability: If a random variable $X$ follows the Binomial distribution $B(33,p)$ such that $3P(X=0)=P(X=1)$, then the value of $\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}$ is equal to

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of arithmetic.

You see a Binomial distribution $X\sim B(33,p)$ and a target expression involving probabilities of 15, 16, 17, and 18 successes. Your instinct might be to panic, to reach for a calculator, or to start writing out massive factorials.

But stop. Take a breath. In the world of JEE Advanced, whenever you see large numbers in a probability expression, there is almost always a hidden symmetry waiting to be exploited. Let us embark on this journey together.

Decoding the Condition

We are given the condition $3P(X=0)=P(X=1)$. Let us translate this into the language of mathematics using the Binomial probability formula:

For $n=33$, the probability of zero successes is $P(X=0)=\left(\genfrac{}{}{0ex}{}{33}{0}\right)p^0{q}^{33}$, which simplifies beautifully to $q^{33}$ because $\left(\genfrac{}{}{0ex}{}{33}{0}\right)=1$ and $p^0=1$.

Similarly, $P(X=1)=\left(\genfrac{}{}{0ex}{}{33}{1}\right)p^1{q}^{32}=33p{q}^{32}$. Substituting these into our given condition, we get $3q^{33}=33p{q}^{32}$.

By dividing both sides by $3q^{32}$ (which is safe because $q$ cannot be zero), we arrive at the elegant result: $q=11p$, or $\frac{q}{p}=11$. Keep this value safe; it is the key that will unlock the entire problem.

The Symmetry Insight

Now, look at the target expression:

Notice the indices: $15+18=33$ and $16+17=33$. This is not a coincidence. This is the geometry of the Binomial distribution.

The coefficients $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ are symmetric around the center. Because $n=33$, the distribution is symmetric around $16.5$. This means $\left(\genfrac{}{}{0ex}{}{33}{15}\right)=\left(\genfrac{}{}{0ex}{}{33}{18}\right)$ and $\left(\genfrac{}{}{0ex}{}{33}{16}\right)=\left(\genfrac{}{}{0ex}{}{33}{17}\right)$.

When we write out the ratio $\frac{P(X=15)}{P(X=18)}$, the binomial coefficients cancel out completely! We are left with:

Similarly, the second term $\frac{P(X=16)}{P(X=17)}$ simplifies to $\frac{q}{p}$.

The Algebraic Collapse

We have reduced a terrifying expression into a simple algebraic one:

We already know that $\frac{q}{p}=11$. Substituting this, we get $11^3-11$.

Calculating this is straightforward: $1331-11=1320$.

Just like that, the complexity melts away. This problem teaches us that in physics and mathematics, brute force is rarely the answer. The answer lies in observing the structure, respecting the symmetry, and simplifying before you calculate. You have mastered the logic; the final answer is 1320.