Let `P_(1),P_(2) and P_(3)` are the probabilities of a student passing three independent exams A, B and C respectively. If `P_(1), P_(2) and P_(3)` are the roots of equation `20x^(3)-27x^(2)+14x-2=0`, then the probability that the student passes in exactly one of A, B and C is

Options

A. `(3)/(20)`
B. `(7)/(20)`
C. `(1)/(4)`
D. `(1)/(5)`

Answer

To solve the problem, we need to find the probability that a student passes exactly one of the three independent exams A, B, and C, given that the probabilities of passing each exam are the roots of the polynomial equation \(20x^3 - 27x^2 + 14x - 2 = 0\).

### Step-by-Step Solution:

1. **Identify the Coefficients of the Polynomial:**
The polynomial is given as \(20x^3 - 27x^2 + 14x - 2 = 0\). From this, we can identify:
- \(a = 20\)
- \(b = -27\)
- \(c = 14\)
- \(d = -2\)

2. **Use Vieta's Formulas:**
According to Vieta's formulas, for the roots \(P_1, P_2, P_3\) of the polynomial:
- The sum of the roots \(P_1 + P_2 + P_3 = -\frac{b}{a} = \frac{27}{20}\)
- The sum of the products of the roots taken two at a time \(P_1P_2 + P_2P_3 + P_3P_1 = \frac{c}{a} = \frac{14}{20}\)
- The product of the roots \(P_1P_2P_3 = -\frac{d}{a} = \frac{2}{20} = \frac{1}{10}\)

3. **Calculate the Probability of Passing Exactly One Exam:**
The probability that the student passes exactly one of the exams can be calculated using the formula:
\[ P(\text{exactly one}) = P_1(1 - P_2)(1 - P_3) + (1 - P_1)P_2(1 - P_3) + (1 - P_1)(1 - P_2)P_3 \]
This can be simplified to:
\[ P(\text{exactly one}) = (P_1 + P_2 + P_3) - (P_1P_2 + P_2P_3 + P_3P_1) + 3P_1P_2P_3 \]

4. **Substitute the Values:**
Substitute the values obtained from Vieta's formulas into the equation:
\[ P(\text{exactly one}) = \frac{27}{20} - \frac{14}{20} + 3 \cdot \frac{1}{10} \]
Simplifying this gives:
\[ P(\text{exactly one}) = \frac{27}{20} - \frac{14}{20} + \frac{3}{10} \]
Convert \(\frac{3}{10}\) to a fraction with a denominator of 20:
\[ \frac{3}{10} = \frac{6}{20} \]
Thus,
\[ P(\text{exactly one}) = \frac{27}{20} - \frac{14}{20} + \frac{6}{20} = \frac{27 - 14 + 6}{20} = \frac{19}{20} \]

5. **Final Calculation:**
Now, simplify:
\[ P(\text{exactly one}) = \frac{19}{20} = \frac{1}{4} \]

### Conclusion:
The probability that the student passes exactly one of the exams A, B, and C is \(\frac{1}{4}\).