An unbiased normal coin is tossed n times. Let
`E_(1):` event that both heads and tails are present in n tosses.
`E_(2):` event that the coin shows up heads at most once.
The value of n for which `E_(1) and E_(2)` are independent is ______.

Explanation

To solve the problem, we need to find the value of \( n \) for which the events \( E_1 \) and \( E_2 \) are independent.

### Step 1: Define the events
- \( E_1 \): Event that both heads and tails are present in \( n \) tosses.
- \( E_2 \): Event that the coin shows up heads at most once.

### Step 2: Calculate \( P(E_1) \)
To find \( P(E_1) \), we can use the complement:
- The only cases where both heads and tails are not present are when all tosses are heads or all tosses are tails.

Thus, the probability of getting all heads or all tails is:
\[ P(\text{All Heads}) + P(\text{All Tails}) = \frac{1}{2^n} + \frac{1}{2^n} = \frac{2}{2^n} = \frac{1}{2^{n-1}} \]

Therefore, the probability of event \( E_1 \) is:
\[ P(E_1) = 1 - P(\text{All Heads or All Tails}) = 1 - \frac{1}{2^{n-1}} = 1 - \frac{1}{2^n} - \frac{1}{2^n} = 1 - \frac{1}{2^{n-1}} \]

### Step 3: Calculate \( P(E_2) \)
Event \( E_2 \) occurs if there are either no heads or exactly one head:
- Probability of no heads (all tails): \( P(\text{No Heads}) = \frac{1}{2^n} \)
- Probability of exactly one head: The number of ways to choose 1 head from \( n \) tosses is \( \binom{n}{1} = n \), and the probability for each arrangement is \( \frac{1}{2^n} \).

Thus, the probability of event \( E_2 \) is:
\[ P(E_2) = P(\text{No Heads}) + P(\text{Exactly One Head}) = \frac{1}{2^n} + n \cdot \frac{1}{2^n} = \frac{1 + n}{2^n} \]

### Step 4: Calculate \( P(E_1 \cap E_2) \)
For \( E_1 \) and \( E_2 \) to occur together, we need to have exactly one head (since having no heads contradicts \( E_1 \)). Therefore:
\[ P(E_1 \cap E_2) = P(E_2) = \frac{1 + n}{2^n} \]

### Step 5: Use the independence condition
Events \( E_1 \) and \( E_2 \) are independent if:
\[ P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) \]

Substituting the probabilities we calculated:
\[ \frac{1 + n}{2^n} = \left(1 - \frac{1}{2^{n-1}}\right) \cdot \frac{1 + n}{2^n} \]

### Step 6: Simplify the equation
Expanding the right side:
\[ \frac{1 + n}{2^n} = \frac{(1 + n) \left(1 - \frac{1}{2^{n-1}}\right)}{2^n} \]
This simplifies to:
\[ 1 + n = (1 + n) \left(1 - \frac{1}{2^{n-1}}\right) \]

### Step 7: Solve for \( n \)
Expanding the right side:
\[ 1 + n = (1 + n) - \frac{(1 + n)}{2^{n-1}} \]
Rearranging gives:
\[ \frac{(1 + n)}{2^{n-1}} = 0 \]
This implies \( n + 1 = 2^{n-1} \).

### Step 8: Find integer solutions
Testing small values of \( n \):
- For \( n = 1 \): \( 1 + 1 = 2 \) (true)
- For \( n = 2 \): \( 2 + 1 = 3 \) (false)
- For \( n = 3 \): \( 3 + 1 = 4 \) (true)

Thus, the only integer solution is \( n = 3 \).

### Final Answer
The value of \( n \) for which \( E_1 \) and \( E_2 \) are independent is \( n = 3 \).