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
- \( E_1 \): The event that both heads and tails are present in \( n \) tosses.
- \( E_2 \): The event that the coin shows heads at most once in \( n \) tosses.

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

The probability of getting all heads in \( n \) tosses is:
\[ P(\text{All Heads}) = \left(\frac{1}{2}\right)^n \]
The probability of getting all tails is the same:
\[ P(\text{All Tails}) = \left(\frac{1}{2}\right)^n \]
Thus, the probability of \( E_1 \) is:
\[ P(E_1) = 1 - P(\text{All Heads}) - P(\text{All Tails}) = 1 - \left(\frac{1}{2}\right)^n - \left(\frac{1}{2}\right)^n = 1 - 2\left(\frac{1}{2}\right)^n = 1 - \frac{2}{2^n} \]

### Step 3: Calculate \( P(E_2) \)
The event \( E_2 \) occurs if we have either 0 heads or 1 head in \( n \) tosses.

- The probability of getting 0 heads (all tails) is:
\[ P(0 \text{ heads}) = \left(\frac{1}{2}\right)^n \]
- The probability of getting exactly 1 head is given by the binomial coefficient:
\[ P(1 \text{ head}) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \left(\frac{1}{2}\right)^n \]
Thus, the probability of \( E_2 \) is:
\[ P(E_2) = P(0 \text{ heads}) + P(1 \text{ head}) = \left(\frac{1}{2}\right)^n + n \left(\frac{1}{2}\right)^n = (n + 1) \left(\frac{1}{2}\right)^n \]

### Step 4: Independence Condition
Events \( E_1 \) and \( E_2 \) are independent if:
\[ P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) \]

### Step 5: Calculate \( P(E_1 \cap E_2) \)
For \( E_2 \) to occur (at most one head), \( E_1 \) can only occur if there is exactly 1 head (since having 0 heads means tails only, which does not satisfy \( E_1 \)). Therefore:
\[ P(E_1 \cap E_2) = P(1 \text{ head}) = n \left(\frac{1}{2}\right)^n \]

### Step 6: Set Up the Equation
Now we set up the equation for independence:
\[ n \left(\frac{1}{2}\right)^n = \left(1 - \frac{2}{2^n}\right) \cdot \left((n + 1) \left(\frac{1}{2}\right)^n\right) \]

### Step 7: Simplify the Equation
This simplifies to:
\[ n = (n + 1) \left(1 - \frac{2}{2^n}\right) \]
Expanding this gives:
\[ n = n + 1 - \frac{2(n + 1)}{2^n} \]
Rearranging leads to:
\[ \frac{2(n + 1)}{2^n} = 1 \]
This implies:
\[ 2(n + 1) = 2^n \]
or
\[ n + 1 = 2^{n-1} \]

### Step 8: Solve for \( n \)
We can test 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)
- For \( n = 4 \): \( 4 + 1 = 5 \) (False)
- For \( n = 5 \): \( 5 + 1 = 6 \) (False)
- For \( n = 6 \): \( 6 + 1 = 7 \) (False)

After testing, we find that \( n = 3 \) is the only solution that satisfies the equation.

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

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