The mean of the 'Sixes' in two tosses of an unbiased die is

Options

A. `(1)/(3)`
B. `(1)/(6)`
C. `(2)/(3)`
D. `(5)/(18)`

Answer

To find the mean of 'Sixes' in two tosses of an unbiased die, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of times a '6' appears when rolling a die twice. The possible values of \( X \) are 0, 1, and 2.

### Step 2: Calculate the Probabilities
We need to find the probabilities for each value of \( X \):

1. **Probability of getting 0 Sixes (X = 0)**:
- This occurs when neither of the two rolls results in a '6'.
- The probability of not rolling a '6' in one roll is \( \frac{5}{6} \).
- Therefore, the probability of getting 0 Sixes in 2 rolls is:
\[ P(X = 0) = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \]

2. **Probability of getting 1 Six (X = 1)**:
- This occurs when one of the rolls results in a '6' and the other does not.
- The number of ways to choose which roll is a '6' is \( \binom{2}{1} = 2 \).
- The probability of this happening is:
\[ P(X = 1) = 2 \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right) = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{10}{36} \]

3. **Probability of getting 2 Sixes (X = 2)**:
- This occurs when both rolls result in a '6'.
- The probability of this happening is:
\[ P(X = 2) = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \]

### Step 3: Summarize the Probabilities
Now we summarize the probabilities:
- \( P(X = 0) = \frac{25}{36} \)
- \( P(X = 1) = \frac{10}{36} \)
- \( P(X = 2) = \frac{1}{36} \)

### Step 4: Calculate the Mean
The mean \( E(X) \) can be calculated using the formula:
\[ E(X) = \sum (x_i \cdot P(X = x_i)) \]
Substituting the values:
\[ E(X) = 0 \cdot \frac{25}{36} + 1 \cdot \frac{10}{36} + 2 \cdot \frac{1}{36} \]
Calculating this gives:
\[ E(X) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3} \]

### Final Answer
The mean of 'Sixes' in two tosses of an unbiased die is \( \frac{1}{3} \).

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