How many times a fair coin must be tossed, so that the probabillity of getting at least one tail is more than 90 %

Answer

To solve the problem of how many times a fair coin must be tossed so that the probability of getting at least one tail is more than 90%, we can follow these steps:

### Step 1: Understand the Probability of Getting Tails
When a fair coin is tossed, the probability of getting heads (H) is \( P(H) = \frac{1}{2} \) and the probability of getting tails (T) is \( P(T) = \frac{1}{2} \).

### Step 2: Calculate the Probability of Getting All Heads
If the coin is tossed \( n \) times, the probability of getting all heads is:
\[ P(\text{All Heads}) = \left(\frac{1}{2}\right)^n \]

### Step 3: Calculate the Probability of Getting At Least One Tail
The probability of getting at least one tail is the complement of getting all heads:
\[ P(\text{At least one Tail}) = 1 - P(\text{All Heads}) = 1 - \left(\frac{1}{2}\right)^n \]

### Step 4: Set Up the Inequality
We want this probability to be greater than 90%, or 0.9:
\[ 1 - \left(\frac{1}{2}\right)^n > 0.9 \]

### Step 5: Simplify the Inequality
Rearranging the inequality gives:
\[ \left(\frac{1}{2}\right)^n < 0.1 \]

### Step 6: Take the Logarithm
To solve for \( n \), we can take logarithms on both sides. Using base 10 logarithm:
\[ \log\left(\left(\frac{1}{2}\right)^n\right) < \log(0.1) \]
This simplifies to:
\[ n \cdot \log\left(\frac{1}{2}\right) < \log(0.1) \]

### Step 7: Substitute Values
We know that \( \log\left(\frac{1}{2}\right) = -\log(2) \) and \( \log(0.1) = -1 \):
\[ n \cdot (-\log(2)) < -1 \]
This can be rewritten as:
\[ n \cdot \log(2) > 1 \]

### Step 8: Solve for \( n \)
Now, we can solve for \( n \):
\[ n > \frac{1}{\log(2)} \]
Using the approximate value \( \log(2) \approx 0.301 \):
\[ n > \frac{1}{0.301} \approx 3.32 \]

### Step 9: Round Up
Since \( n \) must be a whole number (you can't toss a coin a fraction of a time), we round up to the next whole number:
\[ n \geq 4 \]

### Final Answer
Therefore, the minimum number of times a fair coin must be tossed to ensure the probability of getting at least one tail is more than 90% is:
\[ \boxed{4} \]