A die is rolled twice. Getting a number greater than 4 is called a 'Success'. Let the random variable X is equal to the number of successes then mean of X is

Options

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

Answer

To find the mean of the random variable \( X \), which represents the number of successes when a die is rolled twice, we can follow these steps:

### Step 1: Define Success and Failure
A success is defined as rolling a number greater than 4. The successful outcomes when rolling a die are 5 and 6. Therefore, the probability of success \( P(S) \) is:
\[ P(S) = \frac{2}{6} = \frac{1}{3} \]
The probability of failure \( P(F) \) (rolling a number 1, 2, 3, or 4) is:
\[ P(F) = 1 - P(S) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 2: Identify Possible Values of \( X \)
The random variable \( X \) can take the values 0, 1, or 2:
- \( X = 0 \): No successes (both rolls are failures)
- \( X = 1 \): One success (one roll is a success, the other is a failure)
- \( X = 2 \): Two successes (both rolls are successes)

### Step 3: Calculate the Probabilities for Each Value of \( X \)
1. **Probability that \( X = 0 \)**:
\[ P(X = 0) = P(F) \times P(F) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

2. **Probability that \( X = 1 \)**:
\[ P(X = 1) = P(S) \times P(F) + P(F) \times P(S) = 2 \times \left(\frac{1}{3} \times \frac{2}{3}\right) = 2 \times \frac{2}{9} = \frac{4}{9} \]

3. **Probability that \( X = 2 \)**:
\[ P(X = 2) = P(S) \times P(S) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

### Step 4: Calculate the Mean (Expected Value) of \( X \)
The mean (or expected value) \( E(X) \) is calculated using the formula:
\[ E(X) = \sum (x \cdot P(X = x)) \]
Substituting the values we calculated:
\[ E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) \]
\[ E(X) = 0 \cdot \frac{4}{9} + 1 \cdot \frac{4}{9} + 2 \cdot \frac{1}{9} \]
\[ E(X) = 0 + \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3} \]

### Final Answer
The mean of \( X \) is \( \frac{2}{3} \).

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