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 step by step, we need to analyze the events and their probabilities.

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

### Step 2: Calculate the Probability of Event \( E \)
Since the coin is fair, the probability of getting heads on the first flip is:
\[ P(E) = \frac{1}{2} \]

### Step 3: Calculate the Probability of Event \( F(k) \)
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:
\[ P(F(k)) = \frac{\binom{n}{k}}{2^n} \]

### Step 4: Calculate the Probability of Intersection \( P(E \cap F(k)) \)
For the intersection \( E \cap F(k) \), we need to consider that we have a head on the first flip and exactly \( k \) heads in total. If the first flip is a head, 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 is:
\[ P(E \cap F(k)) = \frac{\binom{n-1}{k-1}}{2^n} \]

### Step 5: Set Up the 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:
\[ \frac{\binom{n-1}{k-1}}{2^n} = \frac{1}{2} \cdot \frac{\binom{n}{k}}{2^n} \]

### Step 6: Simplify the Equation
Cancelling \( \frac{1}{2^n} \) from both sides gives:
\[ \binom{n-1}{k-1} = \frac{1}{2} \binom{n}{k} \]

### Step 7: Use the Binomial Coefficient Identity
Using the identity \( \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \), we can rewrite:
\[ \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \]
Thus:
\[ \binom{n-1}{k-1} = \frac{1}{2} \left( \binom{n-1}{k} + \binom{n-1}{k-1} \right) \]

### Step 8: Rearranging the Equation
Multiplying both sides by 2 gives:
\[ 2 \binom{n-1}{k-1} = \binom{n-1}{k} + \binom{n-1}{k-1} \]
This simplifies to:
\[ \binom{n-1}{k-1} = \binom{n-1}{k} \]
This equality holds when \( k = 1 \) or \( n = 2k \).

### Step 9: Find \( \frac{n}{k} \)
From \( n = 2k \), we can find:
\[ \frac{n}{k} = 2 \]

### Final Answer
The value of \( \frac{n}{k} \) for which \( E \) and \( F(k) \) are independent is:
\[ \boxed{2} \]