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 number of times \( n \) a coin is tossed such that the probability of getting at least one head is greater than the probability of getting at least two tails by \( \frac{5}{32} \).

### Step-by-Step Solution:

1. **Define the Probabilities**:
- Let \( P(A) \) be the probability of getting at least one head.
- Let \( P(B) \) be the probability of getting at least two tails.

2. **Calculate \( P(A) \)**:
- The probability of getting at least one head can be calculated using the complement rule:
\[ P(A) = 1 - P(\text{no heads}) \]
- The probability of getting no heads (i.e., all tails) when tossing a coin \( n \) times is:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]
- Therefore:
\[ P(A) = 1 - \left(\frac{1}{2}\right)^n \]

3. **Calculate \( P(B) \)**:
- The probability of getting at least two tails can also be calculated using the complement rule:
\[ P(B) = 1 - P(\text{no tails}) - P(\text{exactly one tail}) \]
- The probability of getting no tails (i.e., all heads) is:
\[ P(\text{no tails}) = \left(\frac{1}{2}\right)^n \]
- The probability of getting exactly one tail is given by:
\[ P(\text{exactly one tail}) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \left(\frac{1}{2}\right)^n \]
- Therefore:
\[ P(B) = 1 - \left(\frac{1}{2}\right)^n - n \left(\frac{1}{2}\right)^n \]
\[ P(B) = 1 - (n + 1) \left(\frac{1}{2}\right)^n \]

4. **Set Up the Inequality**:
- According to the problem, we have:
\[ P(A) - P(B) = \frac{5}{32} \]
- Substituting the expressions for \( P(A) \) and \( P(B) \):
\[ \left(1 - \left(\frac{1}{2}\right)^n\right) - \left(1 - (n + 1) \left(\frac{1}{2}\right)^n\right) = \frac{5}{32} \]
- Simplifying this gives:
\[ -\left(\frac{1}{2}\right)^n + (n + 1) \left(\frac{1}{2}\right)^n = \frac{5}{32} \]
\[ n \left(\frac{1}{2}\right)^n = \frac{5}{32} \]

5. **Solve for \( n \)**:
- Rearranging gives:
\[ n = \frac{5 \cdot 2^n}{32} \]
- Testing integer values for \( n \):
- For \( n = 5 \):
\[ 5 = \frac{5 \cdot 2^5}{32} \Rightarrow 5 = \frac{5 \cdot 32}{32} \Rightarrow 5 = 5 \quad \text{(True)} \]

Thus, the value of \( n \) is \( 5 \).

### Final Answer:
\[ \boxed{5} \]