A coin is tossed n times . If the probability of getting at least one head is greater than 0.9 then the minimum vaue of n is

Options

A. 3
B. 5
C. 4
D. 2

Answer

To solve the problem, we need to find the minimum value of \( n \) such that the probability of getting at least one head when a coin is tossed \( n \) times is greater than \( 0.9 \).

### Step-by-Step Solution:

1. **Understanding the Probability of Getting Heads**:
The probability of getting heads in a single toss of a fair coin is \( \frac{1}{2} \). Consequently, the probability of getting tails (no heads) in a single toss is also \( \frac{1}{2} \).

2. **Probability of Getting No Heads in n Tosses**:
If the coin is tossed \( n \) times, the probability of getting tails (no heads) in all \( n \) tosses is:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]

3. **Probability of Getting At Least One Head**:
The probability of getting at least one head in \( n \) tosses is the complement of getting no heads:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\frac{1}{2}\right)^n \]

4. **Setting Up the Inequality**:
We need this probability to be greater than \( 0.9 \):
\[ 1 - \left(\frac{1}{2}\right)^n > 0.9 \]

5. **Rearranging the Inequality**:
Rearranging gives us:
\[ \left(\frac{1}{2}\right)^n < 0.1 \]

6. **Taking Logarithms**:
To solve for \( n \), we can take logarithms:
\[ n \log\left(\frac{1}{2}\right) < \log(0.1) \]
Since \( \log\left(\frac{1}{2}\right) \) is negative, we can reverse the inequality when dividing:
\[ n > \frac{\log(0.1)}{\log\left(\frac{1}{2}\right)} \]

7. **Calculating the Logarithms**:
We know that:
\[ \log(0.1) = -1 \quad \text{(since } 0.1 = 10^{-1}\text{)} \]
and
\[ \log\left(\frac{1}{2}\right) = -\log(2) \approx -0.3010 \]
Therefore:
\[ n > \frac{-1}{-0.3010} \approx 3.32 \]

8. **Finding the Minimum Integer Value of n**:
Since \( n \) must be a whole number, we take the smallest integer greater than \( 3.32 \), which is \( 4 \).

### Conclusion:
Thus, the minimum value of \( n \) such that the probability of getting at least one head is greater than \( 0.9 \) is:
\[ \boxed{4} \]