A die is tossed twice. 'Getting an odd number on a toss' is considered a success. Find the probability distribution of number of successes. Also, find the mean and variance of the number of successes.

Explanation

To solve the problem step by step, we need to find the probability distribution of the number of successes when a die is tossed twice, where a success is defined as getting an odd number on a toss. We will also calculate the mean and variance of the number of successes.

### Step 1: Identify the Sample Space
When a die is tossed twice, the sample space consists of all possible outcomes. Each die has 6 faces, so the total number of outcomes when tossing two dice is \(6 \times 6 = 36\).

The sample space can be represented as:
\[ S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \]

### Step 2: Define Success
A success is defined as getting an odd number on a toss. The odd numbers on a die are 1, 3, and 5.

### Step 3: Determine the Number of Successes
Let \(X\) be the random variable representing the number of successes (odd numbers) in two tosses. The possible values of \(X\) are:
- \(X = 0\): No odd numbers (both tosses result in even numbers)
- \(X = 1\): One odd number (one toss results in an odd number, and the other in an even number)
- \(X = 2\): Two odd numbers (both tosses result in odd numbers)

### Step 4: Calculate Probabilities
1. **Probability of \(X = 0\)**:
- The even numbers on a die are 2, 4, and 6 (3 even numbers).
- The probability of getting an even number on one toss is \(\frac{3}{6} = \frac{1}{2}\).
- Therefore, the probability of getting two even numbers (0 successes) is:
\[ P(X = 0) = P(\text{even}) \times P(\text{even}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \]

2. **Probability of \(X = 1\)**:
- There are two scenarios for getting one odd number: (odd, even) or (even, odd).
- The probability for each scenario is:
\[ P(X = 1) = P(\text{odd}) \times P(\text{even}) + P(\text{even}) \times P(\text{odd}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

3. **Probability of \(X = 2\)**:
- The probability of getting two odd numbers is:
\[ P(X = 2) = P(\text{odd}) \times P(\text{odd}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \]

### Step 5: Probability Distribution Table
Now we can summarize the probability distribution of \(X\):
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{1}{4} \\ 1 & \frac{1}{2} \\ 2 & \frac{1}{4} \\ \hline \end{array} \]

### Step 6: Calculate the Mean
The mean (expected value) \(E(X)\) is calculated as follows:
\[ E(X) = \sum (x_i \cdot P(X = x_i)) = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} \]
\[ E(X) = 0 + \frac{1}{2} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 7: Calculate the Variance
The variance \(Var(X)\) is calculated using the formula:
\[ Var(X) = E(X^2) - (E(X))^2 \]
First, we calculate \(E(X^2)\):
\[ E(X^2) = \sum (x_i^2 \cdot P(X = x_i)) = 0^2 \cdot \frac{1}{4} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{1}{4} \]
\[ E(X^2) = 0 + \frac{1}{2} + \frac{4}{4} = \frac{1}{2} + 1 = \frac{3}{2} \]
Now, substituting into the variance formula:
\[ Var(X) = \frac{3}{2} - (1)^2 = \frac{3}{2} - 1 = \frac{1}{2} \]

### Final Results
- Probability Distribution:
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{1}{4} \\ 1 & \frac{1}{2} \\ 2 & \frac{1}{4} \\ \hline \end{array} \]
- Mean \(E(X) = 1\)
- Variance \(Var(X) = \frac{1}{2}\)