In a class of 125 students 70 passed in Mathematics, 55 in statistics, and 30 in both. Then find the probability that a student selected at random from the class has passes in only one subject.

Explanation

To solve the problem step by step, we will use the principle of inclusion-exclusion to find the number of students who passed in only one subject (Mathematics or Statistics).

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( A \) be the event that a student passed in Mathematics.
- Let \( B \) be the event that a student passed in Statistics.

2. **Given Data**:
- Total number of students, \( n = 125 \)
- Number of students who passed in Mathematics, \( |A| = 70 \)
- Number of students who passed in Statistics, \( |B| = 55 \)
- Number of students who passed in both subjects, \( |A \cap B| = 30 \)

3. **Calculate Students Passing Only Mathematics**:
- Students who passed only Mathematics can be calculated as:
\[ |A \text{ only}| = |A| - |A \cap B| = 70 - 30 = 40 \]

4. **Calculate Students Passing Only Statistics**:
- Students who passed only Statistics can be calculated as:
\[ |B \text{ only}| = |B| - |A \cap B| = 55 - 30 = 25 \]

5. **Total Students Passing Only One Subject**:
- The total number of students passing only one subject is:
\[ |A \text{ only}| + |B \text{ only}| = 40 + 25 = 65 \]

6. **Calculate the Probability**:
- The probability that a student selected at random passes in only one subject is given by:
\[ P(\text{only one subject}) = \frac{|A \text{ only}| + |B \text{ only}|}{n} = \frac{65}{125} \]
- Simplifying this fraction:
\[ P(\text{only one subject}) = \frac{65 \div 5}{125 \div 5} = \frac{13}{25} \]

### Final Answer:
The probability that a student selected at random from the class has passed in only one subject is \( \frac{13}{25} \).