If the probability mass function of a discrete random variable X is
`P(x)=C/(x^3), x=1,2,3`
=0, otherwise
Then `E(X)=`

Options

A. `343/297`
B. `294/251`
C. `297/294`
D. `251/294`

Answer

To solve the problem, we need to find the expected value \( E(X) \) of the discrete random variable \( X \) given its probability mass function (PMF):

\[ P(X = x) = \frac{C}{x^3}, \quad x = 1, 2, 3 \]
and \( P(X = x) = 0 \) otherwise.

### Step 1: Find the value of \( C \)

To find the value of \( C \), we use the property that the total probability must equal 1:

\[ P(1) + P(2) + P(3) = 1 \]

Substituting the PMF into this equation:

\[ \frac{C}{1^3} + \frac{C}{2^3} + \frac{C}{3^3} = 1 \]

This simplifies to:

\[ C \left( 1 + \frac{1}{8} + \frac{1}{27} \right) = 1 \]

### Step 2: Calculate the sum inside the parentheses

To calculate \( 1 + \frac{1}{8} + \frac{1}{27} \), we find a common denominator. The least common multiple of 1, 8, and 27 is 216.

Calculating each term:

- \( 1 = \frac{216}{216} \)
- \( \frac{1}{8} = \frac{27}{216} \)
- \( \frac{1}{27} = \frac{8}{216} \)

Now, adding these fractions:

\[ 1 + \frac{1}{8} + \frac{1}{27} = \frac{216 + 27 + 8}{216} = \frac{251}{216} \]

### Step 3: Solve for \( C \)

Now substituting back into the equation for \( C \):

\[ C \cdot \frac{251}{216} = 1 \]

Thus,

\[ C = \frac{216}{251} \]

### Step 4: Calculate the expected value \( E(X) \)

The expected value \( E(X) \) is calculated using the formula:

\[ E(X) = \sum_{x=1}^{3} x \cdot P(X = x) \]

Substituting the PMF:

\[ E(X) = 1 \cdot P(1) + 2 \cdot P(2) + 3 \cdot P(3) \]

This becomes:

\[ E(X) = 1 \cdot \frac{C}{1^3} + 2 \cdot \frac{C}{2^3} + 3 \cdot \frac{C}{3^3} \]

Substituting \( C = \frac{216}{251} \):

\[ E(X) = 1 \cdot \frac{216}{251} + 2 \cdot \frac{216}{251 \cdot 8} + 3 \cdot \frac{216}{251 \cdot 27} \]

This simplifies to:

\[ E(X) = \frac{216}{251} \left( 1 + \frac{2}{8} + \frac{3}{27} \right) \]

### Step 5: Calculate the sum inside the parentheses

Calculating \( 1 + \frac{2}{8} + \frac{3}{27} \):

- \( 1 = \frac{216}{216} \)
- \( \frac{2}{8} = \frac{54}{216} \)
- \( \frac{3}{27} = \frac{24}{216} \)

Adding these fractions:

\[ 1 + \frac{2}{8} + \frac{3}{27} = \frac{216 + 54 + 24}{216} = \frac{294}{216} \]

### Step 6: Final calculation of \( E(X) \)

Now substituting back into the expected value:

\[ E(X) = \frac{216}{251} \cdot \frac{294}{216} \]

The \( 216 \) cancels out:

\[ E(X) = \frac{294}{251} \]

Thus, the expected value \( E(X) \) is:

\[ \boxed{\frac{294}{251}} \]