The least number of times a fair coin must be tossed so that the probability of getting atleast one head is atleast 0.8 is

Options

A. 6
B. 5
C. 4
D. 3

Answer

To solve the problem of finding the least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, we can follow these steps:

### Step 1: Understand the Problem
We need to find the minimum number of tosses \( n \) such that the probability of getting at least one head is at least 0.8.

### Step 2: Define the Probability
The probability of getting at least one head when tossing a coin \( n \) times can be expressed as:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) \]
Where \( P(\text{no heads}) \) is the probability of getting tails in all \( n \) tosses.

### Step 3: Calculate the Probability of No Heads
The probability of getting tails in one toss is \( \frac{1}{2} \). Therefore, the probability of getting tails in \( n \) tosses is:
\[ P(\text{no heads}) = \left(\frac{1}{2}\right)^n \]

### Step 4: Set Up the Inequality
We want the probability of getting at least one head to be at least 0.8:
\[ 1 - \left(\frac{1}{2}\right)^n \geq 0.8 \]

### Step 5: Solve the Inequality
Rearranging the inequality gives:
\[ \left(\frac{1}{2}\right)^n \leq 0.2 \]

### Step 6: Convert to Exponential Form
Taking the reciprocal, we get:
\[ 2^n \geq \frac{1}{0.2} \]
Calculating \( \frac{1}{0.2} \):
\[ \frac{1}{0.2} = 5 \]
Thus, we have:
\[ 2^n \geq 5 \]

### Step 7: Find the Minimum \( n \)
Now we need to find the smallest integer \( n \) such that \( 2^n \geq 5 \):
- For \( n = 1 \): \( 2^1 = 2 \) (not sufficient)
- For \( n = 2 \): \( 2^2 = 4 \) (not sufficient)
- For \( n = 3 \): \( 2^3 = 8 \) (sufficient)

Thus, the least number of times the coin must be tossed is \( n = 3 \).

### Final Answer
The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8 is **3**.

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