Suppose `E_1, E_2 and E_3` be three mutually exclusive events such that `P(E_i)=p_i" for "i=1, 2, 3`.
If `p_1, p_2 and p_3 ` are the roots of `27x^3-27x^2+ax-1=0` the value of a is

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem, we need to find the value of \( a \) given that \( p_1, p_2, \) and \( p_3 \) are the roots of the polynomial equation \( 27x^3 - 27x^2 + ax - 1 = 0 \).

### Step-by-step Solution:

1. **Identify the coefficients of the polynomial:**
The polynomial is given as:
\[ 27x^3 - 27x^2 + ax - 1 = 0 \]
Here, the coefficients are:
- \( a = 27 \) (coefficient of \( x^3 \))
- \( b = -27 \) (coefficient of \( x^2 \))
- \( c = a \) (coefficient of \( x \))
- \( d = -1 \) (constant term)

2. **Use Vieta's Formulas:**
According to Vieta's formulas for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( p_1 + p_2 + p_3 = -\frac{b}{a} \)
- The sum of the products of the roots taken two at a time \( p_1p_2 + p_2p_3 + p_1p_3 = \frac{c}{a} \)
- The product of the roots \( p_1p_2p_3 = -\frac{d}{a} \)

3. **Calculate the sum of the roots:**
From Vieta's, we have:
\[ p_1 + p_2 + p_3 = -\frac{-27}{27} = 1 \]

4. **Calculate the product of the roots:**
The product of the roots is given by:
\[ p_1p_2p_3 = -\frac{-1}{27} = \frac{1}{27} \]

5. **Use the Arithmetic Mean-Geometric Mean Inequality:**
Since \( p_1, p_2, \) and \( p_3 \) are probabilities and must be non-negative, we apply the AM-GM inequality:
\[ \frac{p_1 + p_2 + p_3}{3} \geq \sqrt[3]{p_1p_2p_3} \]
Substituting the known values:
\[ \frac{1}{3} \geq \sqrt[3]{\frac{1}{27}} \]
Simplifying the right side:
\[ \sqrt[3]{\frac{1}{27}} = \frac{1}{3} \]
Thus, we have:
\[ \frac{1}{3} = \frac{1}{3} \]
This equality holds when \( p_1 = p_2 = p_3 \).

6. **Set the roots equal:**
Since \( p_1 = p_2 = p_3 \), we can denote them as \( p \). Thus:
\[ 3p = 1 \implies p = \frac{1}{3} \]

7. **Calculate the sum of products of the roots taken two at a time:**
We know:
\[ p_1p_2 + p_2p_3 + p_1p_3 = 3p^2 = 3\left(\frac{1}{3}\right)^2 = 3 \cdot \frac{1}{9} = \frac{1}{3} \]

8. **Relate it to \( a \):**
From Vieta's, we have:
\[ p_1p_2 + p_2p_3 + p_1p_3 = \frac{a}{27} \]
Thus:
\[ \frac{1}{3} = \frac{a}{27} \]
Solving for \( a \):
\[ a = 27 \cdot \frac{1}{3} = 9 \]

### Final Answer:
The value of \( a \) is \( \boxed{9} \).