A random variable x had its range {0, 1, 2} and the probabilities are given by `P(x=0)=3k^(3), P(x=1)=4k-10k^(2), P(x=2)=5k-1` where k is constant then k =

Options

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

Answer

To solve the problem, we need to determine the value of the constant \( k \) such that the probabilities \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \) sum to 1. The probabilities are given as follows:

- \( P(X=0) = 3k^3 \)
- \( P(X=1) = 4k - 10k^2 \)
- \( P(X=2) = 5k - 1 \)

### Step-by-Step Solution:

1. **Set up the equation for the sum of probabilities:**
\[ P(X=0) + P(X=1) + P(X=2) = 1 \]
Substituting the expressions for the probabilities:
\[ 3k^3 + (4k - 10k^2) + (5k - 1) = 1 \]

2. **Combine like terms:**
\[ 3k^3 - 10k^2 + 4k + 5k - 1 = 1 \]
Simplifying this gives:
\[ 3k^3 - 10k^2 + 9k - 1 = 1 \]

3. **Rearrange the equation:**
\[ 3k^3 - 10k^2 + 9k - 2 = 0 \]

4. **Check for possible rational roots using the Rational Root Theorem:**
We can test \( k = 1 \):
\[ 3(1)^3 - 10(1)^2 + 9(1) - 2 = 3 - 10 + 9 - 2 = 0 \]
Since \( k = 1 \) is a root, we can factor the polynomial.

5. **Factor the polynomial:**
Since \( k - 1 \) is a factor, we can perform polynomial long division or synthetic division to find the other factor:
\[ 3k^3 - 10k^2 + 9k - 2 = (k - 1)(3k^2 - 7k + 2) \]

6. **Solve the quadratic equation:**
Now we need to solve \( 3k^2 - 7k + 2 = 0 \) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]
\[ = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \]
This gives us:
\[ k = \frac{12}{6} = 2 \quad \text{or} \quad k = \frac{2}{6} = \frac{1}{3} \]

7. **Check the validity of the values of \( k \):**
- For \( k = 2 \):
- \( P(X=0) = 3(2)^3 = 24 \) (not a valid probability)
- For \( k = \frac{1}{3} \):
- \( P(X=0) = 3\left(\frac{1}{3}\right)^3 = 3 \cdot \frac{1}{27} = \frac{1}{9} \)
- \( P(X=1) = 4\left(\frac{1}{3}\right) - 10\left(\frac{1}{3}\right)^2 = \frac{4}{3} - \frac{10}{9} = \frac{12}{9} - \frac{10}{9} = \frac{2}{9} \)
- \( P(X=2) = 5\left(\frac{1}{3}\right) - 1 = \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \)

- Check if these probabilities sum to 1:
\[ \frac{1}{9} + \frac{2}{9} + \frac{6}{9} = \frac{1 + 2 + 6}{9} = \frac{9}{9} = 1 \]

### Conclusion:
The valid value of \( k \) is \( \frac{1}{3} \).