A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads, is

Options

A. `15//2^8`
B. `2//15`
C. `15//2^13`
D. none of these

Explanation

To solve the problem, we need to find the probability of getting two heads when a fair coin is tossed a fixed number of times, given that the probability of getting seven heads is equal to the probability of getting nine heads.

### Step-by-Step Solution:

1. **Understanding the Probability of Heads**:
- When a fair coin is tossed, the probability of getting heads (H) is \( p = \frac{1}{2} \) and the probability of getting tails (T) is \( q = 1 - p = \frac{1}{2} \).

2. **Using Binomial Probability Formula**:
- The probability of getting exactly \( k \) heads in \( n \) tosses is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- For our case, we have:
\[ P(X = 7) = \binom{n}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{n-7} = \binom{n}{7} \left(\frac{1}{2}\right)^n \]
\[ P(X = 9) = \binom{n}{9} \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{n-9} = \binom{n}{9} \left(\frac{1}{2}\right)^n \]

3. **Setting the Probabilities Equal**:
- Since the probability of getting seven heads is equal to the probability of getting nine heads, we can set the two equations equal:
\[ \binom{n}{7} \left(\frac{1}{2}\right)^n = \binom{n}{9} \left(\frac{1}{2}\right)^n \]
- We can cancel \( \left(\frac{1}{2}\right)^n \) from both sides (as long as \( n \) is not zero):
\[ \binom{n}{7} = \binom{n}{9} \]

4. **Using the Property of Binomial Coefficients**:
- The property of binomial coefficients states that \( \binom{n}{k} = \binom{n}{n-k} \). Thus:
\[ \binom{n}{9} = \binom{n}{n-9} \]
- Therefore, we have:
\[ \binom{n}{7} = \binom{n}{n-9} \]

5. **Finding the Value of \( n \)**:
- For \( \binom{n}{7} = \binom{n}{n-9} \), we can set \( n - 9 = 7 \):
\[ n = 16 \]

6. **Calculating the Probability of Getting Two Heads**:
- Now, we need to find the probability of getting exactly two heads when the coin is tossed 16 times:
\[ P(X = 2) = \binom{16}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{16-2} = \binom{16}{2} \left(\frac{1}{2}\right)^{16} \]
- Calculate \( \binom{16}{2} \):
\[ \binom{16}{2} = \frac{16 \times 15}{2 \times 1} = 120 \]
- Therefore:
\[ P(X = 2) = 120 \left(\frac{1}{2}\right)^{16} = 120 \times \frac{1}{65536} = \frac{120}{65536} = \frac{15}{8192} \]

### Final Answer:
The probability of getting two heads when a fair coin is tossed 16 times is \( \frac{15}{8192} \).