The probability to pass in an examination of mathematics for three students are `1/4,1/5,1/6`. Find the probability that at least two students will pass in this examination.

Explanation

To find the probability that at least two students will pass the examination, we can use the complementary probability approach. We will first calculate the probabilities of the complementary events (i.e., fewer than two students passing) and then subtract from 1.

### Step 1: Define the probabilities
Let:
- \( P(A) = \frac{1}{4} \) (Probability that Student A passes)
- \( P(B) = \frac{1}{5} \) (Probability that Student B passes)
- \( P(C) = \frac{1}{6} \) (Probability that Student C passes)

The probabilities that each student fails are:
- \( P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4} \)
- \( P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5} \)
- \( P(C') = 1 - P(C) = 1 - \frac{1}{6} = \frac{5}{6} \)

### Step 2: Calculate the probability of fewer than 2 students passing
The events for fewer than 2 students passing include:
1. No students pass.
2. Exactly one student passes.

#### Case 1: No students pass
The probability that none of the students pass is:
\[ P(A') \times P(B') \times P(C') = \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \]
Calculating this:
\[ = \frac{3 \times 4 \times 5}{4 \times 5 \times 6} = \frac{3}{6} = \frac{1}{2} \]

#### Case 2: Exactly one student passes
We need to calculate the probabilities for each student passing while the others fail.

- Probability that only Student A passes:
\[ P(A) \times P(B') \times P(C') = \frac{1}{4} \times \frac{4}{5} \times \frac{5}{6} = \frac{1 \times 4 \times 5}{4 \times 5 \times 6} = \frac{1}{6} \]

- Probability that only Student B passes:
\[ P(A') \times P(B) \times P(C') = \frac{3}{4} \times \frac{1}{5} \times \frac{5}{6} = \frac{3 \times 1 \times 5}{4 \times 5 \times 6} = \frac{3}{24} = \frac{1}{8} \]

- Probability that only Student C passes:
\[ P(A') \times P(B') \times P(C) = \frac{3}{4} \times \frac{4}{5} \times \frac{1}{6} = \frac{3 \times 4 \times 1}{4 \times 5 \times 6} = \frac{3}{30} = \frac{1}{10} \]

Now, we sum these probabilities for exactly one student passing:
\[ P(\text{Exactly 1 passes}) = \frac{1}{6} + \frac{1}{8} + \frac{1}{10} \]
Finding a common denominator (which is 120):
\[ = \frac{20}{120} + \frac{15}{120} + \frac{12}{120} = \frac{47}{120} \]

### Step 3: Total probability of fewer than 2 students passing
Now, we add the probabilities of the two cases:
\[ P(\text{Fewer than 2 pass}) = P(\text{No pass}) + P(\text{Exactly 1 pass}) = \frac{1}{2} + \frac{47}{120} \]
Converting \(\frac{1}{2}\) to a fraction with a denominator of 120:
\[ \frac{1}{2} = \frac{60}{120} \]
Thus,
\[ P(\text{Fewer than 2 pass}) = \frac{60}{120} + \frac{47}{120} = \frac{107}{120} \]

### Step 4: Probability that at least 2 students pass
Finally, we find the probability that at least 2 students pass:
\[ P(\text{At least 2 pass}) = 1 - P(\text{Fewer than 2 pass}) = 1 - \frac{107}{120} = \frac{120 - 107}{120} = \frac{13}{120} \]

### Final Answer
The probability that at least two students will pass the examination is:
\[ \frac{13}{120} \]