(i) A problem in Mathematics is given to three students whose chances of solving it are :
`(1)/(2),(1)/(4)` and `(1)/(5)`.
What is the probability that at least one of them may solve it ?
(ii) A problem is given to three students, whose chances of solving it are :
`(1)/(3),(1)/(5)` and `(1)/(6)`.
What is the probability that exactly one of them may solve it.

Answer

To solve the given problems step by step, let's break them down:

### Part (i)

**Problem Statement:** A problem in Mathematics is given to three students whose chances of solving it are \( \frac{1}{2}, \frac{1}{4}, \) and \( \frac{1}{5} \). What is the probability that at least one of them may solve it?

**Step 1:** Identify the probabilities of each student solving the problem.
- Let \( P(A) = \frac{1}{2} \) (probability that student A solves it)
- Let \( P(B) = \frac{1}{4} \) (probability that student B solves it)
- Let \( P(C) = \frac{1}{5} \) (probability that student C solves it)

**Step 2:** Calculate the probabilities that each student does NOT solve the problem.
- \( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)
- \( P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} \)
- \( P(C') = 1 - P(C) = 1 - \frac{1}{5} = \frac{4}{5} \)

**Step 3:** Calculate the probability that none of the students solve the problem.
- \( P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') \)
- \( P(A' \cap B' \cap C') = \frac{1}{2} \times \frac{3}{4} \times \frac{4}{5} \)

**Step 4:** Perform the multiplication.
- \( P(A' \cap B' \cap C') = \frac{1}{2} \times \frac{3}{4} \times \frac{4}{5} = \frac{1 \times 3 \times 4}{2 \times 4 \times 5} = \frac{12}{40} = \frac{3}{10} \)

**Step 5:** Calculate the probability that at least one student solves the problem.
- \( P(\text{at least one solves}) = 1 - P(A' \cap B' \cap C') \)
- \( P(\text{at least one solves}) = 1 - \frac{3}{10} = \frac{7}{10} \)

**Final Answer for Part (i):** The probability that at least one of them may solve it is \( \frac{7}{10} \).

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### Part (ii)

**Problem Statement:** A problem is given to three students, whose chances of solving it are \( \frac{1}{3}, \frac{1}{5}, \) and \( \frac{1}{6} \). What is the probability that exactly one of them may solve it?

**Step 1:** Identify the probabilities of each student solving the problem.
- Let \( P(A) = \frac{1}{3} \)
- Let \( P(B) = \frac{1}{5} \)
- Let \( P(C) = \frac{1}{6} \)

**Step 2:** Calculate the probabilities that each student does NOT solve the problem.
- \( P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \)
- \( P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5} \)
- \( P(C') = 1 - P(C) = 1 - \frac{1}{6} = \frac{5}{6} \)

**Step 3:** Calculate the probability that exactly one student solves the problem. This can happen in three different scenarios:
1. A solves it, B and C do not.
2. B solves it, A and C do not.
3. C solves it, A and B do not.

**Step 4:** Calculate the probabilities for each scenario.
- **Scenario 1:** \( P(A) \times P(B') \times P(C') = \frac{1}{3} \times \frac{4}{5} \times \frac{5}{6} = \frac{20}{90} \)
- **Scenario 2:** \( P(A') \times P(B) \times P(C') = \frac{2}{3} \times \frac{1}{5} \times \frac{5}{6} = \frac{10}{90} \)
- **Scenario 3:** \( P(A') \times P(B') \times P(C) = \frac{2}{3} \times \frac{4}{5} \times \frac{1}{6} = \frac{8}{90} \)

**Step 5:** Add the probabilities of all three scenarios to find the total probability of exactly one student solving the problem.
- Total Probability = \( \frac{20}{90} + \frac{10}{90} + \frac{8}{90} = \frac{38}{90} = \frac{19}{45} \)

**Final Answer for Part (ii):** The probability that exactly one of them may solve it is \( \frac{19}{45} \).

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