In order to get atleast once a head with probability P `ge `0.9. the number of times a coin needs to be tossed is :

Options

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

Answer

To solve the problem of determining how many times a coin needs to be tossed to achieve at least one head with a probability of at least 0.9, we can follow these steps:

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

**Hint:** Remember that the total probability of all outcomes must equal 1.

### Step 2: Calculate the Probability of Getting No Heads
If we toss the coin \( n \) times, the probability of getting no heads (i.e., getting tails every time) is given by:
\[ P(\text{no heads in } n \text{ tosses}) = P(T)^n = \left(\frac{1}{2}\right)^n \]

**Hint:** Think about what it means to not get a head in multiple tosses.

### Step 3: Calculate the Probability of Getting At Least One Head
The probability of getting at least one head in \( n \) tosses is the complement of getting no heads:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\frac{1}{2}\right)^n \]

**Hint:** The complement rule is useful here; it helps simplify the calculation.

### Step 4: Set Up the Inequality
We want the probability of getting at least one head to be at least 0.9:
\[ 1 - \left(\frac{1}{2}\right)^n \geq 0.9 \]

**Hint:** Rearranging the inequality will help isolate \( n \).

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

Taking the logarithm of both sides (base 2 for simplicity):
\[ -n \log_2(2) \leq \log_2(0.1) \]
\[ -n \leq \log_2(0.1) \]
\[ n \geq -\log_2(0.1) \]

**Hint:** Remember that \( \log_2(0.1) \) is a negative number, which will flip the inequality when multiplied by -1.

### Step 6: Calculate \( \log_2(0.1) \)
Using the change of base formula:
\[ \log_2(0.1) = \frac{\log_{10}(0.1)}{\log_{10}(2)} \approx \frac{-1}{0.301} \approx -3.32 \]
Thus,
\[ n \geq 3.32 \]

Since \( n \) must be a whole number, we round up to the nearest whole number:
\[ n \geq 4 \]

**Hint:** Always round up when dealing with probabilities in practical scenarios.

### Conclusion
The minimum number of times the coin needs to be tossed to achieve at least a 0.9 probability of getting at least one head is \( n = 4 \).

**Final Answer:** The number of times a coin needs to be tossed is **4**.