A fair coin is flipped n times. Let E be the event "a head is obtained on the first flip" and let `F_(k)` be the event "exactly k heads are obtained". Then the value of n/k for which E and `F_(k)` are independent is _____.

Explanation

To solve the problem, we need to find the value of \( \frac{n}{k} \) for which the events \( E \) (a head on the first flip) and \( F_k \) (exactly \( k \) heads in \( n \) flips) are independent.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( E \) be the event that a head is obtained on the first flip.
- Let \( F_k \) be the event that exactly \( k \) heads are obtained in \( n \) flips.

2. **Calculate \( P(E) \)**:
- Since the coin is fair, the probability of getting a head on the first flip is:
\[ P(E) = \frac{1}{2} \]

3. **Calculate \( P(F_k) \)**:
- The total number of outcomes when flipping a coin \( n \) times is \( 2^n \).
- The number of ways to get exactly \( k \) heads in \( n \) flips is given by the binomial coefficient \( \binom{n}{k} \).
- Therefore, the probability of getting exactly \( k \) heads is:
\[ P(F_k) = \binom{n}{k} \left(\frac{1}{2}\right)^n \]

4. **Calculate \( P(E \cap F_k) \)**:
- For \( E \) and \( F_k \) to occur together, we must have a head on the first flip and exactly \( k \) heads in total. This means we need \( k-1 \) heads from the remaining \( n-1 \) flips.
- The number of ways to choose \( k-1 \) heads from \( n-1 \) flips is \( \binom{n-1}{k-1} \).
- Thus, the probability of both \( E \) and \( F_k \) occurring is:
\[ P(E \cap F_k) = \binom{n-1}{k-1} \left(\frac{1}{2}\right)^n \]

5. **Independence Condition**:
- Events \( E \) and \( F_k \) are independent if:
\[ P(E \cap F_k) = P(E) \cdot P(F_k) \]
- Substituting the probabilities we calculated:
\[ \binom{n-1}{k-1} \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right) \cdot \left(\binom{n}{k} \left(\frac{1}{2}\right)^n\right) \]
- Simplifying this gives:
\[ \binom{n-1}{k-1} = \frac{1}{2} \cdot \binom{n}{k} \]

6. **Using the Binomial Coefficient Identity**:
- We know that:
\[ \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \]
- Therefore, substituting this into our equation:
\[ \binom{n-1}{k-1} = \frac{1}{2} \left( \binom{n-1}{k} + \binom{n-1}{k-1} \right) \]
- Rearranging gives:
\[ 2 \binom{n-1}{k-1} = \binom{n-1}{k} + \binom{n-1}{k-1} \]

7. **Solving for \( n \) and \( k \)**:
- This leads to the conclusion that \( n = 2k \).

8. **Finding \( \frac{n}{k} \)**:
- Therefore, we can find:
\[ \frac{n}{k} = \frac{2k}{k} = 2 \]

### Final Answer:
The value of \( \frac{n}{k} \) for which \( E \) and \( F_k \) are independent is \( 2 \).