The minimum number of times one has to toss a fair coin so that the probability of observing atleast one head is atlest `90%` is

Options

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

Answer

To solve the problem of finding the minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We want the probability of getting at least one head when tossing a fair coin \( n \) times to be at least 90%.

2. **Defining the Probability**:
The probability of getting at least one head in \( n \) tosses can be calculated using the complement rule:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) \]
For a fair coin, the probability of getting tails (no heads) in one toss is \( \frac{1}{2} \).

3. **Calculating the Probability of No Heads**:
The probability of getting no heads in \( n \) tosses (i.e., getting tails every time) is:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]

4. **Setting Up the Inequality**:
We want:
\[ 1 - \left(\frac{1}{2}\right)^n \geq 0.9 \]
Rearranging this gives:
\[ \left(\frac{1}{2}\right)^n \leq 0.1 \]

5. **Solving the Inequality**:
To solve for \( n \), we can take logarithms:
\[ n \log\left(\frac{1}{2}\right) \leq \log(0.1) \]
Since \( \log\left(\frac{1}{2}\right) \) is negative, we can multiply both sides by -1, which reverses the inequality:
\[ n \geq \frac{\log(0.1)}{\log\left(\frac{1}{2}\right)} \]

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

7. **Finding the Minimum Integer Value**:
Since \( n \) must be a whole number, we round up to the nearest whole number:
\[ n = 4 \]

Thus, the minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is **4**.