A sample space consists of 3 sample points with associated probabilities given as `2p ,p^2,4p-1.` Then the value of `p` is

Options

A. `p = sqrt(11) - 3`
B. `sqrt(10) - 3`
C. `(1)/(4)lt p lt (1)/(2)`
D. none of these

Explanation

To solve the problem, we need to find the value of \( p \) given that the probabilities associated with three sample points are \( 2p \), \( p^2 \), and \( 4p - 1 \). The sum of these probabilities must equal 1, as they represent the entire sample space.

### Step-by-Step Solution:

1. **Set up the equation for total probability:**
\[ 2p + p^2 + (4p - 1) = 1 \]

2. **Combine like terms:**
\[ 2p + p^2 + 4p - 1 = 1 \]
This simplifies to:
\[ p^2 + 6p - 1 = 1 \]

3. **Rearrange the equation:**
\[ p^2 + 6p - 2 = 0 \]

4. **Identify coefficients for the quadratic formula:**
Here, \( a = 1 \), \( b = 6 \), and \( c = -2 \).

5. **Apply the quadratic formula:**
The quadratic formula is given by:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ p = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]

6. **Calculate the discriminant:**
\[ p = \frac{-6 \pm \sqrt{36 + 8}}{2} \]
\[ p = \frac{-6 \pm \sqrt{44}}{2} \]

7. **Simplify the square root:**
\[ \sqrt{44} = \sqrt{4 \cdot 11} = 2\sqrt{11} \]
Thus, we have:
\[ p = \frac{-6 \pm 2\sqrt{11}}{2} \]

8. **Simplify further:**
\[ p = -3 \pm \sqrt{11} \]

9. **Determine the valid solution:**
We have two potential solutions:
\[ p = -3 + \sqrt{11} \quad \text{and} \quad p = -3 - \sqrt{11} \]
Since probabilities must be non-negative, we only consider \( p = -3 + \sqrt{11} \).

### Conclusion:
The value of \( p \) is:
\[ p = -3 + \sqrt{11} \]