A coin is tossed n times. The probability of getting at least one head is greater than that of getting at least two tails by 5/32. Then n is :

Options

A. 5
B. 10
C. 15
D. None of these

Answer

To solve the problem, we need to find the value of \( n \) such that the probability of getting at least one head when a coin is tossed \( n \) times is greater than the probability of getting at least two tails by \( \frac{5}{32} \).

### Step-by-Step Solution:

1. **Understanding the Probabilities**:
- The probability of getting at least one head when a coin is tossed \( n \) times can be calculated as:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - P(0 \text{ heads}) \]
- The probability of getting no heads (which means getting all tails) is:
\[ P(0 \text{ heads}) = \left(\frac{1}{2}\right)^n \]
- Therefore, we have:
\[ P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^n \]

2. **Calculating the Probability of Getting at Least Two Tails**:
- The probability of getting at least two tails can be calculated as:
\[ P(\text{at least two tails}) = 1 - P(0 \text{ tails}) - P(1 \text{ tail}) \]
- The probability of getting 0 tails (which means getting all heads) is:
\[ P(0 \text{ tails}) = \left(\frac{1}{2}\right)^n \]
- The probability of getting exactly 1 tail is:
\[ P(1 \text{ tail}) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \left(\frac{1}{2}\right)^n \]
- Therefore, we have:
\[ P(\text{at least two tails}) = 1 - \left(\frac{1}{2}\right)^n - n \left(\frac{1}{2}\right)^n \]
- This simplifies to:
\[ P(\text{at least two tails}) = 1 - (n + 1) \left(\frac{1}{2}\right)^n \]

3. **Setting Up the Equation**:
- According to the problem, the difference between the probabilities is given by:
\[ P(\text{at least one head}) - P(\text{at least two tails}) = \frac{5}{32} \]
- Substituting the probabilities we derived:
\[ \left(1 - \left(\frac{1}{2}\right)^n\right) - \left(1 - (n + 1) \left(\frac{1}{2}\right)^n\right) = \frac{5}{32} \]
- This simplifies to:
\[ (n + 1) \left(\frac{1}{2}\right)^n - \left(\frac{1}{2}\right)^n = \frac{5}{32} \]
- Thus, we have:
\[ n \left(\frac{1}{2}\right)^n = \frac{5}{32} \]

4. **Solving for \( n \)**:
- Rearranging gives:
\[ n = \frac{5 \cdot 2^n}{32} \]
- Testing integer values for \( n \):
- For \( n = 5 \):
\[ 5 = \frac{5 \cdot 2^5}{32} \implies 5 = \frac{5 \cdot 32}{32} \implies 5 = 5 \quad \text{(True)} \]
- Thus, \( n = 5 \) satisfies the equation.

### Final Answer:
The value of \( n \) is \( 5 \).