A random variable X takes values 0, 1, 2, 3,… with probability proportional to `(x+1)((1)/(5))^x`.
`P(Xge2)` equals

Options

A. `(11)/(25)`
B. `(13)/(25)`
C. `(11)/(125)`
D. `(13)/(125)`

Answer

To solve the problem, we need to find \( P(X \geq 2) \) for the random variable \( X \) which takes values \( 0, 1, 2, 3, \ldots \) with probabilities proportional to \( (x+1) \left( \frac{1}{5} \right)^x \).

### Step-by-Step Solution:

1. **Define the Probability Function**:
The probability \( P(X = x) \) is proportional to \( (x + 1) \left( \frac{1}{5} \right)^x \). We can express this as:
\[ P(X = x) = k (x + 1) \left( \frac{1}{5} \right)^x \]
where \( k \) is a normalization constant.

2. **Find the Normalization Constant \( k \)**:
To find \( k \), we need to ensure that the total probability sums to 1:
\[ \sum_{x=0}^{\infty} P(X = x) = 1 \]
This gives us:
\[ \sum_{x=0}^{\infty} k (x + 1) \left( \frac{1}{5} \right)^x = 1 \]
We can factor out \( k \):
\[ k \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = 1 \]

3. **Calculate the Sum**:
The sum \( \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x \) can be split into two parts:
\[ \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = \sum_{x=0}^{\infty} x \left( \frac{1}{5} \right)^x + \sum_{x=0}^{\infty} \left( \frac{1}{5} \right)^x \]
The second sum is a geometric series:
\[ \sum_{x=0}^{\infty} \left( \frac{1}{5} \right)^x = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4} \]
The first sum can be calculated using the formula for the expected value of a geometric distribution:
\[ \sum_{x=0}^{\infty} x r^x = \frac{r}{(1 - r)^2} \quad \text{for } r = \frac{1}{5} \]
Thus,
\[ \sum_{x=0}^{\infty} x \left( \frac{1}{5} \right)^x = \frac{\frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{1}{5} \cdot \frac{25}{16} = \frac{5}{16} \]
Therefore,
\[ \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = \frac{5}{16} + \frac{5}{4} = \frac{5}{16} + \frac{20}{16} = \frac{25}{16} \]

4. **Solve for \( k \)**:
Now substituting back:
\[ k \cdot \frac{25}{16} = 1 \implies k = \frac{16}{25} \]

5. **Calculate \( P(X \geq 2) \)**:
We can find \( P(X \geq 2) \) using:
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]
First, calculate \( P(X = 0) \):
\[ P(X = 0) = k (0 + 1) \left( \frac{1}{5} \right)^0 = \frac{16}{25} \cdot 1 = \frac{16}{25} \]
Next, calculate \( P(X = 1) \):
\[ P(X = 1) = k (1 + 1) \left( \frac{1}{5} \right)^1 = \frac{16}{25} \cdot 2 \cdot \frac{1}{5} = \frac{32}{125} \]
Now substituting into \( P(X \geq 2) \):
\[ P(X \geq 2) = 1 - \frac{16}{25} - \frac{32}{125} \]
Convert \( \frac{16}{25} \) to a fraction with a denominator of 125:
\[ \frac{16}{25} = \frac{80}{125} \]
Thus,
\[ P(X \geq 2) = 1 - \frac{80}{125} - \frac{32}{125} = 1 - \frac{112}{125} = \frac{13}{125} \]

### Final Answer:
\[ P(X \geq 2) = \frac{13}{125} \]