The probabilities of a student getting I, II and III division in an examination are `(1)/(10),(3)/(5)` and `(1)/(4)\ ` respectively. The probability that the student fails in the examination is

Options

A. a) `(197)/(200)`
B. b) `(27)/(100)`
C. c) `(83)/(100)`
D. d) none of these

Explanation

To solve the problem, we first need to understand the probabilities given for the divisions and then find the probability of failing the examination.

### Step-by-Step Solution:

1. **Identify the Given Probabilities:**
- Probability of getting I division (P(I)) = \( \frac{1}{10} \)
- Probability of getting II division (P(II)) = \( \frac{3}{5} \)
- Probability of getting III division (P(III)) = \( \frac{1}{4} \)

2. **Calculate the Total Probability of Getting a Division:**
To find the probability of failing, we first need to calculate the total probability of getting any division (I, II, or III). This is done by adding the probabilities of getting each division:
\[ P(I) + P(II) + P(III) = \frac{1}{10} + \frac{3}{5} + \frac{1}{4} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 10, 5, and 4 is 20.

- Convert each probability to have a denominator of 20:
\[ P(I) = \frac{1}{10} = \frac{2}{20} \]
\[ P(II) = \frac{3}{5} = \frac{12}{20} \]
\[ P(III) = \frac{1}{4} = \frac{5}{20} \]

- Now add them:
\[ P(I) + P(II) + P(III) = \frac{2}{20} + \frac{12}{20} + \frac{5}{20} = \frac{19}{20} \]

3. **Calculate the Probability of Failing:**
The probability of failing (P(Fail)) is the complement of the probability of getting any division:
\[ P(Fail) = 1 - (P(I) + P(II) + P(III)) \]
\[ P(Fail) = 1 - \frac{19}{20} = \frac{1}{20} \]

### Final Answer:
The probability that the student fails in the examination is \( \frac{1}{20} \).

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