The probabilities of a student getting first class or second class or third class in an examination are `(2)/(7) , (3)/(5) ,(1)/(10)` respectively . The probability that the students fails is .

Options

A. `(6)/(70)`
B. `(11)/(70)`
C. `(3)/(35)`
D. `(1)/(70)`

Answer

To solve the problem, we need to find the probability that a student fails the examination given the probabilities of passing with different classifications.

### Step-by-Step Solution:

1. **Identify the given probabilities:**
- Probability of getting 1st class, \( P(A) = \frac{2}{7} \)
- Probability of getting 2nd class, \( P(B) = \frac{3}{5} \)
- Probability of getting 3rd class, \( P(C) = \frac{1}{10} \)

2. **Set up the equation for total probability:**
The total probability of all possible outcomes (getting 1st class, 2nd class, 3rd class, or failing) must equal 1. Therefore, we can write:
\[ P(A) + P(B) + P(C) + P(F) = 1 \]
where \( P(F) \) is the probability of failing.

3. **Substitute the known probabilities into the equation:**
\[ \frac{2}{7} + \frac{3}{5} + \frac{1}{10} + P(F) = 1 \]

4. **Find a common denominator:**
The least common multiple (LCM) of the denominators 7, 5, and 10 is 70. We will convert each fraction to have a denominator of 70:
- \( P(A) = \frac{2}{7} = \frac{2 \times 10}{7 \times 10} = \frac{20}{70} \)
- \( P(B) = \frac{3}{5} = \frac{3 \times 14}{5 \times 14} = \frac{42}{70} \)
- \( P(C) = \frac{1}{10} = \frac{1 \times 7}{10 \times 7} = \frac{7}{70} \)

5. **Substitute these values back into the equation:**
\[ \frac{20}{70} + \frac{42}{70} + \frac{7}{70} + P(F) = 1 \]

6. **Combine the fractions:**
\[ \frac{20 + 42 + 7}{70} + P(F) = 1 \]
\[ \frac{69}{70} + P(F) = 1 \]

7. **Solve for \( P(F) \):**
\[ P(F) = 1 - \frac{69}{70} \]
\[ P(F) = \frac{70}{70} - \frac{69}{70} = \frac{1}{70} \]

### Final Answer:
The probability that the student fails is \( \frac{1}{70} \).