Three students appear at an examination of mathematics. The probability of their success are `1/3,1/4,1/5` respectively. Find the probability of success of at least two.

Answer

To find the probability of at least two students succeeding in the examination, we can follow these steps:

### Step 1: Define the probabilities of success and failure for each student.
- Let the probability of success for student S1 be \( P(S1) = \frac{1}{3} \).
- Let the probability of success for student S2 be \( P(S2) = \frac{1}{4} \).
- Let the probability of success for student S3 be \( P(S3) = \frac{1}{5} \).

Now, we can calculate the probabilities of failure for each student:
- Probability of failure for S1:
\[ P(F1) = 1 - P(S1) = 1 - \frac{1}{3} = \frac{2}{3} \]
- Probability of failure for S2:
\[ P(F2) = 1 - P(S2) = 1 - \frac{1}{4} = \frac{3}{4} \]
- Probability of failure for S3:
\[ P(F3) = 1 - P(S3) = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 2: Identify the scenarios for at least two successes.
The scenarios for at least two students succeeding are:
1. Exactly two students succeed.
2. All three students succeed.

### Step 3: Calculate the probability for each scenario.

#### Case 1: Exactly two students succeed.
- **S1 and S2 succeed, S3 fails:**
\[ P(S1) \cdot P(S2) \cdot P(F3) = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{4}{5} = \frac{1 \cdot 1 \cdot 4}{3 \cdot 4 \cdot 5} = \frac{4}{60} \]

- **S1 and S3 succeed, S2 fails:**
\[ P(S1) \cdot P(F2) \cdot P(S3) = \frac{1}{3} \cdot \frac{3}{4} \cdot \frac{1}{5} = \frac{1 \cdot 3 \cdot 1}{3 \cdot 4 \cdot 5} = \frac{3}{60} \]

- **S2 and S3 succeed, S1 fails:**
\[ P(F1) \cdot P(S2) \cdot P(S3) = \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} = \frac{2 \cdot 1 \cdot 1}{3 \cdot 4 \cdot 5} = \frac{2}{60} \]

#### Case 2: All three students succeed.
- **All succeed:**
\[ P(S1) \cdot P(S2) \cdot P(S3) = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} = \frac{1 \cdot 1 \cdot 1}{3 \cdot 4 \cdot 5} = \frac{1}{60} \]

### Step 4: Sum the probabilities from both cases.
Now, we add the probabilities from all scenarios:
\[ P(\text{at least 2 successes}) = \left(\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}\right) = \frac{10}{60} = \frac{1}{6} \]

### Final Answer:
The probability of at least two students succeeding is \( \frac{1}{6} \).

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