A person gets Rs. `(2X+3)` where X denotes the number appearing when a balaced die is rolled once, how much money can be expect in the long run per game ?

Answer

To solve the problem, we need to find the expected amount of money a person can expect to receive when rolling a balanced die and receiving an amount of Rs. `(2X + 3)`, where `X` is the outcome of the die roll.

### Step-by-Step Solution:

1. **Identify the Possible Outcomes:**
The possible outcomes when rolling a balanced die are:
\[ X = 1, 2, 3, 4, 5, 6 \]

2. **Determine the Probability of Each Outcome:**
Since the die is balanced, the probability of each outcome is:
\[ P(X = x) = \frac{1}{6} \quad \text{for } x = 1, 2, 3, 4, 5, 6 \]

3. **Calculate the Expected Value of X:**
The expected value \( E(X) \) is calculated using the formula:
\[ E(X) = \sum (x \cdot P(X = x)) \]
Substituting the values:
\[ E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \]
Simplifying this:
\[ E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2} = 3.5 \]

4. **Calculate the Expected Amount:**
Now, we need to calculate the expected amount the person receives, which is given by \( E(2X + 3) \):
\[ E(2X + 3) = 2E(X) + 3 \]
Substituting \( E(X) \):
\[ E(2X + 3) = 2 \cdot 3.5 + 3 = 7 + 3 = 10 \]

5. **Final Result:**
Therefore, the expected amount of money the person can expect to receive in the long run per game is:
\[ \text{Expected Amount} = \text{Rs. } 10 \]