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 a random variable \( X \) that takes values \( 0, 1, 2, 3, \ldots \) with probabilities proportional to \( (x+1) \left( \frac{1}{5} \right)^x \).

### Step 1: Define the Probability Function
The probability \( P(X = x) \) is proportional to \( (x+1) \left( \frac{1}{5} \right)^x \). Therefore, we can write:
\[ P(X = x) = k (x + 1) \left( \frac{1}{5} \right)^x \]
where \( k \) is a normalization constant.

### Step 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:
\[ \sum_{x=0}^{\infty} k (x + 1) \left( \frac{1}{5} \right)^x = 1 \]
Factoring out \( k \):
\[ k \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = 1 \]

### Step 3: Calculate the Series
We can split the series:
\[ \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = \sum_{x=0}^{\infty} x \left( \frac{1}{5} \right)^x + \sum_{x=0}^{\infty} \left( \frac{1}{5} \right)^x \]
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 sum of an arithmetico-geometric series:
\[ \sum_{x=0}^{\infty} x r^x = \frac{r}{(1 - r)^2} \]
Substituting \( r = \frac{1}{5} \):
\[ \sum_{x=0}^{\infty} x \left( \frac{1}{5} \right)^x = \frac{\frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{1}{5} \cdot \frac{25}{16} = \frac{5}{16} \]

### Step 4: Combine the Results
Now we can combine the results:
\[ \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{5} \right)^x = \frac{5}{16} + \frac{5}{4} = \frac{5}{16} + \frac{20}{16} = \frac{25}{16} \]
Thus, we have:
\[ k \cdot \frac{25}{16} = 1 \implies k = \frac{16}{25} \]

### Step 5: Write the Probability Function
Now we can write:
\[ P(X = x) = \frac{16}{25} (x + 1) \left( \frac{1}{5} \right)^x \]

### Step 6: Calculate \( P(X \geq 2) \)
To find \( P(X \geq 2) \), we can use the complement:
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]
Calculating \( P(X = 0) \) and \( P(X = 1) \):
\[ P(X = 0) = \frac{16}{25} (0 + 1) \left( \frac{1}{5} \right)^0 = \frac{16}{25} \]
\[ P(X = 1) = \frac{16}{25} (1 + 1) \left( \frac{1}{5} \right)^1 = \frac{16}{25} \cdot 2 \cdot \frac{1}{5} = \frac{32}{125} \]

### Step 7: Combine to Find \( P(X \geq 2) \)
Now substituting these values:
\[ P(X \geq 2) = 1 - \frac{16}{25} - \frac{32}{125} \]
Finding a common denominator (125):
\[ P(X \geq 2) = 1 - \frac{80}{125} - \frac{32}{125} = 1 - \frac{112}{125} = \frac{13}{125} \]

### Final Answer
Thus, the final answer is:
\[ P(X \geq 2) = \frac{13}{125} \]