Animated Solution for Mathematics - Probability: A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

Unlocking Binomial Distributions

Welcome, future engineers. Today, we are not just solving a probability problem; we are peeling back the curtain on the elegant, rhythmic nature of the Binomial Distribution.

When you look at a problem like this, do not see it as a dry calculation. See it as a landscape. Imagine a fair coin, tossed $n$ times. Every toss is a binary choice—a fork in the road—leading to either a head or a tail.

When we toss this coin repeatedly, we are tracing the shape of a probability distribution, a curve that rises to a peak and falls away with perfect, mathematical grace.

Defining the Random Variable

Let us ground ourselves in the basics. We define a random variable $X$ as the number of heads obtained in $n$ tosses.

Because the coin is fair, the probability of success (getting a head), denoted by $p$, is $\frac{1}{2}$. Consequently, the probability of failure (getting a tail), denoted by $q$, is also $\frac{1}{2}$.

In any binomial distribution, the probability of getting exactly $r$ successes in $n$ trials is governed by the powerful formula:

This formula is your best friend in the JEE examination hall. It encapsulates the number of ways to arrange $r$ successes (${}^n{C}_r$) multiplied by the probability of those specific outcomes occurring ($p^r{q}^{n-r}$).

The Insight of Symmetry

Now, look at the condition provided: $P(X=7)=P(X=9)$.

If you visualize the graph of a binomial distribution, you will see a bell-like shape. If the probability of getting 7 heads is exactly the same as the probability of getting 9 heads, it implies that the distribution is perfectly balanced between these two points.

Mathematically, we substitute our values into the formula:

Observe the exponents. On the left, we have ${\left(\frac{1}{2}\right)}^7\times {\left(\frac{1}{2}\right)}^{n-7}$, which simplifies to ${\left(\frac{1}{2}\right)}^n$. On the right, we have ${\left(\frac{1}{2}\right)}^9\times {\left(\frac{1}{2}\right)}^{n-9}$, which also simplifies to ${\left(\frac{1}{2}\right)}^n$.

Since ${\left(\frac{1}{2}\right)}^n$ is non-zero, we can confidently cancel it from both sides. We are left with the elegant core of the problem:

The Power of Combinatorial Properties

Here is where the JEE tests your conceptual depth. We know the property of combinations: if ${}^n{C}_x={}^n{C}_y$, then either $x=y$ or $x+y=n$.

Since $7$ is clearly not equal to $9$, the only logical conclusion is that $x+y=n$. Therefore, $7+9=n$, which gives us $n=16$.

We have just discovered that the coin was tossed 16 times. This symmetry is not just a trick; it is the geometric reality of the binomial distribution.

The Final Calculation

Now that we know $n=16$, our target is to find the probability of getting exactly 2 heads, or $P(X=2)$. We return to our trusty formula:

This simplifies to:

Calculating ${}^{16}{C}_2$ is straightforward. Remember the formula ${}^n{C}_r=\frac{n!}{r!(n-r)!}$. For ${}^{16}{C}_2$, this is:

So, our probability is $\frac{120}{2^{16}}$. To match the options provided in the question, we need to simplify this fraction. We can write $120$ as $15\times 8$, and since $8=2^3$, we have:

And there we have it. The answer is $\frac{15}{2^{13}}$, which corresponds to option (1).

Do you see how the complexity melted away once we identified the symmetry? That is the secret to mastering JEE mathematics.

Do not rush into calculations. Pause, visualize the symmetry, use the properties of the functions, and let the algebra do the heavy lifting for you. Keep practicing, keep visualizing, and you will find that these problems are not obstacles—they are stepping stones to your success.