A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of success is

Options

A. `2/9`
B. `4/9`
C. `1/3`
D. `2/3`

Answer

To solve the problem step by step, we need to find the variance of the probability distribution of the number of successes when a die is tossed twice, with success defined as rolling a number greater than 4.

### Step 1: Identify Successes
When a die is rolled, the outcomes are {1, 2, 3, 4, 5, 6}. The numbers greater than 4 are {5, 6}. Thus, the successful outcomes are 5 and 6.

### Step 2: Calculate the Probability of Success
The probability of success (rolling a number greater than 4) is:
- Number of successful outcomes = 2 (5 and 6)
- Total outcomes = 6

So, the probability of success \( P(S) \) is:
\[ P(S) = \frac{2}{6} = \frac{1}{3} \]

### Step 3: Calculate the Probability of Failure
The probability of failure (rolling a number less than or equal to 4) is:
\[ P(S') = 1 - P(S) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 4: Define the Random Variable
Let \( X \) be the random variable representing the number of successes in two tosses of the die. The possible values of \( X \) are 0, 1, or 2.

### Step 5: Calculate the Probabilities for Each Value of \( X \)
1. **Probability of 0 successes (X = 0)**:
\[ P(X = 0) = P(S') \cdot P(S') = \left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right) = \frac{4}{9} \]

2. **Probability of 1 success (X = 1)**:
There are two scenarios for getting exactly one success:
- Success on the first toss and failure on the second: \( P(S) \cdot P(S') = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9} \)
- Failure on the first toss and success on the second: \( P(S') \cdot P(S) = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9} \)

Therefore,
\[ P(X = 1) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9} \]

3. **Probability of 2 successes (X = 2)**:
\[ P(X = 2) = P(S) \cdot P(S) = \left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right) = \frac{1}{9} \]

### Step 6: Summary of Probabilities
- \( P(X = 0) = \frac{4}{9} \)
- \( P(X = 1) = \frac{4}{9} \)
- \( P(X = 2) = \frac{1}{9} \)

### Step 7: Calculate the Expected Value \( E(X) \)
\[ E(X) = \sum (x_i \cdot P(X = x_i)) = 0 \cdot \frac{4}{9} + 1 \cdot \frac{4}{9} + 2 \cdot \frac{1}{9} = 0 + \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3} \]

### Step 8: Calculate \( E(X^2) \)
\[ E(X^2) = \sum (x_i^2 \cdot P(X = x_i)) = 0^2 \cdot \frac{4}{9} + 1^2 \cdot \frac{4}{9} + 2^2 \cdot \frac{1}{9} = 0 + \frac{4}{9} + \frac{4}{9} = \frac{8}{9} \]

### Step 9: Calculate the Variance
The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = E(X^2) - (E(X))^2 \]
Substituting the values:
\[ \sigma^2 = \frac{8}{9} - \left(\frac{2}{3}\right)^2 = \frac{8}{9} - \frac{4}{9} = \frac{4}{9} \]

### Final Answer
The variance of the probability distribution of the number of successes is:
\[ \boxed{\frac{4}{9}} \]