If the probability of A to fail in an examination is `1/5` and that of B is `3/10` , then the probability exactly one of A or B to fail is (i) `11/25` (ii) `19/50` (iii) `1/2` (iv) `3/50`

Options

A. `11/25`
B. `19/50`
C. `1/2`
D. `3/50`

Answer

To solve the problem, we need to find the probability that exactly one of A or B fails in the examination.

1. **Identify the probabilities**:
- The probability that A fails, \( P(A) = \frac{1}{5} \).
- The probability that B fails, \( P(B) = \frac{3}{10} \).

2. **Calculate the probabilities that A and B pass**:
- The probability that A passes, \( P(A') = 1 - P(A) = 1 - \frac{1}{5} = \frac{4}{5} \).
- The probability that B passes, \( P(B') = 1 - P(B) = 1 - \frac{3}{10} = \frac{7}{10} \).

3. **Calculate the probability that exactly one of A or B fails**:
- The event that exactly one of them fails can happen in two ways:
1. A fails and B passes.
2. B fails and A passes.
- The probability that A fails and B passes is:
\[ P(A \text{ fails and } B \text{ passes}) = P(A) \times P(B') = \frac{1}{5} \times \frac{7}{10} = \frac{7}{50}. \]
- The probability that B fails and A passes is:
\[ P(B \text{ fails and } A \text{ passes}) = P(B) \times P(A') = \frac{3}{10} \times \frac{4}{5} = \frac{12}{50}. \]

4. **Combine the probabilities**:
- Now, we add the two probabilities calculated above:
\[ P(\text{exactly one fails}) = P(A \text{ fails and } B \text{ passes}) + P(B \text{ fails and } A \text{ passes}) = \frac{7}{50} + \frac{12}{50} = \frac{19}{50}. \]

5. **Conclusion**:
- Therefore, the probability that exactly one of A or B fails is \( \frac{19}{50} \).

### Final Answer:
The correct option is (ii) \( \frac{19}{50} \).