A coin is tossed any number of times until a head appears. If x donotes the number of tosses till head uappear then `P(x ge 3)=`

Options

A. `(1)/(2)`
B. 1
C. `(1)/(4)`
D. `(3)/(4)`

Answer

To solve the problem of finding \( P(X \geq 3) \), where \( X \) is the number of tosses until the first head appears, we can use the concept of geometric probability.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a fair coin until we get a head. The random variable \( X \) represents the number of tosses until the first head appears. We want to find the probability that it takes 3 or more tosses to get the first head.

2. **Identifying the Probability of Outcomes**:
- The probability of getting a head (H) on any single toss is \( P(H) = \frac{1}{2} \).
- The probability of getting a tail (T) on any single toss is \( P(T) = \frac{1}{2} \).

3. **Calculating \( P(X \geq 3) \)**:
- For \( X \) to be greater than or equal to 3, the first two tosses must be tails. This means we need to get T on the first toss and T on the second toss.
- The probability of getting T on the first toss is \( \frac{1}{2} \) and the probability of getting T on the second toss is also \( \frac{1}{2} \).
- Therefore, the probability of getting T on the first two tosses is:
\[ P(T_1 \cap T_2) = P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

4. **Finding \( P(X \geq 3) \)**:
- Since the event \( X \geq 3 \) occurs if we have tails on the first two tosses, we can conclude:
\[ P(X \geq 3) = P(T_1 \cap T_2) = \frac{1}{4} \]

5. **Final Answer**:
Thus, the probability that it takes 3 or more tosses to get the first head is:
\[ P(X \geq 3) = \frac{1}{4} \]

### Conclusion:
The answer is \( \frac{1}{4} \).

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