Minimum number of times a fair coin must be tossed so that the probility of gettig atleast one head is more than `99%` is

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem, we need to find the minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99%.

### Step-by-step Solution:

1. **Understanding the Probability**:
The probability of getting at least one head when tossing a coin can be expressed as:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) \]
We want this probability to be greater than 99%, or:
\[ P(\text{at least one head}) > 0.99 \]

2. **Setting Up the Inequality**:
From the above expression, we can write:
\[ 1 - P(\text{no heads}) > 0.99 \]
Rearranging gives us:
\[ P(\text{no heads}) < 0.01 \]

3. **Calculating the Probability of No Heads**:
When tossing a fair coin \( n \) times, the probability of getting no heads (i.e., getting tails every time) is:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]
Therefore, we need:
\[ \left(\frac{1}{2}\right)^n < 0.01 \]

4. **Rewriting the Inequality**:
We can rewrite \( 0.01 \) as \( \frac{1}{100} \):
\[ \left(\frac{1}{2}\right)^n < \frac{1}{100} \]
Taking the reciprocal gives:
\[ 2^n > 100 \]

5. **Finding the Minimum \( n \)**:
Now we need to find the smallest integer \( n \) such that \( 2^n > 100 \). We can test values of \( n \):
- For \( n = 6 \): \( 2^6 = 64 \) (not greater than 100)
- For \( n = 7 \): \( 2^7 = 128 \) (greater than 100)

Thus, the minimum \( n \) that satisfies the condition is \( n = 7 \).

### Conclusion:
The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is:
\[ \boxed{7} \]