The number of tosses that have to be made in order that there is `99%` probability of getting at least one head is-

Options

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

Answer

To solve the problem of finding the number of tosses needed to achieve at least a 99% probability of getting at least one head, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the number of coin tosses (n) such that the probability of getting at least one head is at least 99%.

2. **Probability of No Heads**:
The probability of getting no heads in n tosses of a fair coin (which has a probability of 1/2 for heads and 1/2 for tails) is given by:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]

3. **Probability of At Least One Head**:
The probability of getting at least one head 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 want this probability to be at least 99%, which can be expressed mathematically as:
\[ 1 - \left(\frac{1}{2}\right)^n \geq 0.99 \]

5. **Solving the Inequality**:
Rearranging the inequality gives:
\[ \left(\frac{1}{2}\right)^n \leq 0.01 \]

6. **Converting to Logarithmic Form**:
To solve for n, we can take the logarithm of both sides. Using base 2 logarithm:
\[ n \log_2\left(\frac{1}{2}\right) \leq \log_2(0.01) \]
Since \(\log_2\left(\frac{1}{2}\right) = -1\), we can rewrite it as:
\[ -n \leq \log_2(0.01) \]
This simplifies to:
\[ n \geq -\log_2(0.01) \]

7. **Calculating \(-\log_2(0.01)\)**:
We know that:
\[ 0.01 = \frac{1}{100} = 10^{-2} \]
Using the change of base formula:
\[ \log_2(0.01) = \log_2\left(\frac{1}{100}\right) = -\log_2(100) \]
We can approximate \(\log_2(100)\) using:
\[ \log_2(100) \approx 6.644 (since 100 = 10^2 and \log_2(10) \approx 3.32193) \]
Thus:
\[ -\log_2(0.01) \approx 6.644 \]

8. **Finding the Smallest Integer n**:
Since n must be an integer, we round up to the nearest whole number:
\[ n \geq 7 \]

9. **Conclusion**:
Therefore, the minimum number of tosses required to achieve at least a 99% probability of getting at least one head is:
\[ n = 7 \]

### Final Answer:
The number of tosses that have to be made in order that there is a 99% probability of getting at least one head is **7**.