1. Key Concepts and Formulas

• Binomial Probability Distribution: This distribution is used for experiments consisting of a fixed number of independent trials, where each trial has only two possible outcomes (success or failure), and the probability of success remains constant for each trial. The probability of getting exactly $k$ successes in $n$ trials is given by: $P(X=k) = {^nC_k} p^k (1-p)^{n-k}$ where $n$ is the total number of trials, $k$ is the number of successes, $p$ is the probability of success in a single trial, and ${^nC_k} = \frac{n!}{k!(n-k)!}$.
• Fair Coin Condition: For a fair coin, the probability of getting a Head (success), $p$, is $\frac{1}{2}$. Consequently, the probability of getting a Tail (failure), $1-p$, is also $\frac{1}{2}$. Substituting $p = \frac{1}{2}$ into the binomial probability formula simplifies it to: $P(X=k) = {^nC_k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{n-k} = {^nC_k} \left(\frac{1}{2}\right)^{n} = \frac{{^nC_k}}{2^n}$
• Combinatorial Identity: A crucial property of binomial coefficients states that if ${^nC_x} = {^nC_y}$, then either $x=y$ or $x+y=n$. This identity is key to finding the value of $n$ when two binomial probabilities are equal.

2. Step-by-Step Solution

Step 1: Define the Binomial Probability for a Fair Coin Why this step? The problem involves tossing a fair coin a fixed number of times and counting the number of heads. This perfectly fits the Binomial Probability Distribution model. Since the coin is fair, we can simplify the general formula.

Let $n$ be the total number of times the coin is tossed. Let $X$ be the random variable representing the number of heads obtained. For a fair coin, the probability of getting a head is $p = \frac{1}{2}$, and the probability of getting a tail is $1-p = \frac{1}{2}$.

The probability of getting exactly $k$ heads in $n$ tosses is: $P(X=k) = {^nC_k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{n-k}$ Using the property $a^m \cdot a^n = a^{m+n}$, we simplify the powers of $\frac{1}{2}$: $P(X=k) = {^nC_k} \left(\frac{1}{2}\right)^{k + (n-k)} = {^nC_k} \left(\frac{1}{2}\right)^n = \frac{{^nC_k}}{2^n}$ This simplified formula will be used throughout the problem.

Step 2: Use the Given Condition to Determine the Total Number of Tosses ($n$) Why this step? The problem states that the probability of getting 7 heads is equal to the probability of getting 9 heads. This condition allows us to set up an equation to solve for the unknown total number of tosses, $n$.

According to the problem statement: $P(X=7) = P(X=9)$ Using our simplified formula from Step 1: $\frac{{^nC_7}}{2^n} = \frac{{^nC_9}}{2^n}$ Since $2^n$ is a positive non-zero value, we can cancel it from both sides of the equation: ${^nC_7} = {^nC_9}$ Now, we apply the combinatorial identity: if ${^nC_x} = {^nC_y}$, then either $x=y$ or $x+y=n$. Since $7 \neq 9$, we must have the second case: $n = 7 + 9$ $n = 16$ Thus, the fair coin was tossed a total of 16 times.

Step 3: Calculate the Probability of Getting 2 Heads Why this step? Now that we know the total number of tosses ($n=16$), we can answer the main question: find the probability of getting 2 heads.

We need to calculate $P(X=2)$ with $n=16$ and $k=2$. Using the simplified formula from Step 1: $P(X=2) = \frac{{^{16}C_2}}{2^{16}}$ First, calculate the binomial coefficient ${^{16}C_2}$: ${^{16}C_2} = \frac{16!}{2!(16-2)!} = \frac{16!}{2!14!} = \frac{16 \times 15 \times 14!}{ (2 \times 1) \times 14!} = \frac{16 \times 15}{2} = 8 \times 15 = 120$ Now, substitute this value back into the probability expression: $P(X=2) = \frac{120}{2^{16}}$ To match the options, we need to simplify this fraction. We can express $120$ as a product involving powers of 2: $120 = 15 \times 8 = 15 \times 2^3$ Substitute this back into the expression: $P(X=2) = \frac{15 \times 2^3}{2^{16}}$ Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $P(X=2) = \frac{15}{2^{16-3}} = \frac{15}{2^{13}}$

3. Common Mistakes & Tips

• Incorrectly Applying Combinatorial Identity: A common mistake is to forget the $x+y=n$ case when ${^nC_x} = {^nC_y}$, especially when $x \neq y$. Always remember both possibilities.
• Calculation Errors with Powers of 2: Powers of 2 can grow large quickly. Be careful with calculations like $2^{16}$ and simplifying fractions by cancelling powers of 2 (e.g., $120 = 15 \times 2^3$).
• Forgetting Binomial Conditions: Always quickly check if the problem fits the binomial distribution criteria (fixed trials, two outcomes, constant probability, independence) before applying the formula.

4. Summary

We began by establishing the appropriate framework of the Binomial Probability Distribution for a fair coin, simplifying the probability formula to $P(X=k) = \frac{{^nC_k}}{2^n}$. Next, we used the given condition that the probability of getting 7 heads equals the probability of getting 9 heads. By applying the combinatorial identity ${^nC_x} = {^nC_y} \implies x+y=n$, we determined the total number of coin tosses to be $n=16$. Finally, we used this value of $n$ to calculate the probability of getting 2 heads, simplifying the result to $\frac{15}{2^{13}}$.

The final answer is $\boxed{\frac{15}{2^{13}}}$, which corresponds to option (C).