Tossing a coin until a head appear.     Find the sample space,random variance and the probability 

Let's solve this problem step by step:

### 1. **Sample Space (S):**
The problem asks about tossing a coin until a head appears. This implies we keep tossing the coin until we get the first head. The possible outcomes depend on how many tosses it takes for the head to appear. 

- On the first toss, we could get a head (H).
- On the second toss, we could get a tail (T) followed by a head (TH).
- On the third toss, we could get two tails followed by a head (TTH), and so on.

Thus, the sample space \( S \) consists of all sequences where the first head appears at different stages:

\[
S = \{ H, TH, TTH, TTTH, \dots \}
\]

Each outcome in this sample space corresponds to a sequence of coin tosses where tails occur first and the first head appears at a specific toss.

### 2. **Random Variable (X):**
Define a random variable \( X \) to represent the number of tosses needed to get the first head. The possible values for \( X \) are:

\[
X = \{1, 2, 3, 4, \dots \}
\]

This represents that it might take 1 toss (for a head on the first toss), 2 tosses (for a head on the second toss), and so on. The value of \( X \) increases as the number of initial tails increases.

### 3. **Probability Distribution:**
The coin tosses are independent, with each toss having a probability of \( \frac{1}{2} \) for heads (H) and \( \frac{1}{2} \) for tails (T).

The probability of getting a head on the \( n \)-th toss (i.e., the first \( n-1 \) tosses are tails, and the \( n \)-th is a head) follows a geometric distribution. The probability mass function (PMF) for this is:

\[
P(X = n) = (1/2)^{n}
\]

This can be derived as follows:
- The probability of getting \( n-1 \) tails is \( (1/2)^{n-1} \),
- The probability of getting a head on the \( n \)-th toss is \( 1/2 \),
- Therefore, the total probability is \( (1/2)^n \).

### 4. **Expected Value (E[X]):**
The expected value \( E[X] \) for a geometric distribution with parameter \( p = 1/2 \) is given by:

\[
E[X] = \frac{1}{p} = 2
\]

So, on average, it will take 2 tosses to get the first head.

### 5. **Variance (Var[X]):**
The variance for a geometric distribution is given by:

\[
\text{Var}(X) = \frac{1 - p}{p^2} = \frac{1 - \frac{1}{2}}{\left(\frac{1}{2}\right)^2} = 2
\]

Thus, the variance of the number of tosses to get the first head is 2.

### Summary:
- **Sample Space**: \( S = \{H, TH, TTH, TTTH, \dots\} \)
- **Random Variable (X)**: Number of tosses until the first head appears, \( X = \{1, 2, 3, 4, \dots \} \)
- **Probability Distribution**: \( P(X = n) = (1/2)^n \)
- **Expected Value**: \( E[X] = 2 \)
- **Variance**: \( \text{Var}(X) = 2 \)

---

Do you want any more details or have questions? Here are some related questions:

1. How does the variance change with a biased coin (not fair)?
2. What is the probability of getting heads after exactly 3 tosses?
3. How would the distribution change if we stopped after a fixed number of tosses?
4. How would the expected value change if the coin was biased?
5. How do geometric distributions relate to exponential distributions?

**Tip:** For any geometric distribution, the expected value is always \( \frac{1}{p} \), where \( p \) is the probability of success on each trial!

Tossing a coin until a head appears. Find the sample space, random variance, and the probability.

Explore the probability of tossing a coin until a head appears with detailed explanations of the sample space, random variance, and probability distribution using a geometric distribution. Suitable for Grades 9-12.

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