The p.m.f. of a.r.v. X is as follows:
`P( X=0) = 3k^ (3), P(X=1) = 4k-10k^ (2)`, P(X=2)=5k-1, P(X=x)=0` for any other value of x, then value of k is

Options

A. `1/3`
B. `2/3`
C. `1`
D. `2`

Answer

To find the value of \( k \) for the given probability mass function (p.m.f.) of the random variable \( X \), we will follow these steps:

### Step 1: Write down the p.m.f. equations
The p.m.f. is given as:
- \( P(X=0) = 3k^3 \)
- \( P(X=1) = 4k - 10k^2 \)
- \( P(X=2) = 5k - 1 \)
- \( P(X=x) = 0 \) for any other value of \( x \)

### Step 2: Use the property of total probability
The total probability must equal 1. Therefore, we can set up the equation:
\[ P(X=0) + P(X=1) + P(X=2) = 1 \]

Substituting the expressions for the probabilities:
\[ 3k^3 + (4k - 10k^2) + (5k - 1) = 1 \]

### Step 3: Simplify the equation
Combine like terms:
\[ 3k^3 - 10k^2 + 4k + 5k - 1 = 1 \]
\[ 3k^3 - 10k^2 + 9k - 1 = 1 \]
Now, move 1 to the left side:
\[ 3k^3 - 10k^2 + 9k - 2 = 0 \]

### Step 4: Solve the cubic equation
We need to find the roots of the cubic equation:
\[ 3k^3 - 10k^2 + 9k - 2 = 0 \]

Using the trial and error method, we can check for possible rational roots. Testing \( k = 1 \):
\[ 3(1)^3 - 10(1)^2 + 9(1) - 2 = 3 - 10 + 9 - 2 = 0 \]
So, \( k = 1 \) is a root.

### Step 5: Factor the cubic equation
Since \( k = 1 \) is a root, we can factor the cubic polynomial:
\[ 3k^3 - 10k^2 + 9k - 2 = (k - 1)(3k^2 - 7k + 2) \]

### Step 6: Factor the quadratic equation
Now, we will factor the quadratic \( 3k^2 - 7k + 2 \):
\[ 3k^2 - 7k + 2 = 0 \]
Using the factorization method:
\[ 3k^2 - 6k - k + 2 = 0 \]
\[ 3k(k - 2) - 1(k - 2) = 0 \]
\[ (3k - 1)(k - 2) = 0 \]

### Step 7: Find the values of \( k \)
Setting each factor to zero gives us:
1. \( 3k - 1 = 0 \) → \( k = \frac{1}{3} \)
2. \( k - 2 = 0 \) → \( k = 2 \)
3. \( k - 1 = 0 \) → \( k = 1 \)

### Step 8: Check which values are valid
We need to check which of these values keep the probabilities valid (between 0 and 1):
- For \( k = 1 \):
- \( P(X=0) = 3(1)^3 = 3 \) (not valid)
- For \( k = 2 \):
- \( P(X=0) = 3(2)^3 = 24 \) (not valid)
- For \( k = \frac{1}{3} \):
- \( P(X=0) = 3\left(\frac{1}{3}\right)^3 = \frac{3}{27} = \frac{1}{9} \) (valid)

Thus, the only valid value of \( k \) is:
\[ \boxed{\frac{1}{3}} \]