In an examination in a school having 60 students in XII A and 40 students in XII B, 15 and 20 students are failed in XII A and XII B, respectively. A student is called by the principal at random, find the probability that he is from XII A if he passed in the examination.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Gather the given information
- Number of students in XII A = 60
- Number of students in XII B = 40
- Students failed in XII A = 15
- Students failed in XII B = 20

### Step 2: Calculate the number of students who passed
- Students passed in XII A = Total students in XII A - Students failed in XII A
\[ = 60 - 15 = 45 \]
- Students passed in XII B = Total students in XII B - Students failed in XII B
\[ = 40 - 20 = 20 \]

### Step 3: Calculate the total number of students who passed
- Total students passed = Students passed in XII A + Students passed in XII B
\[ = 45 + 20 = 65 \]

### Step 4: Define the events
- Let \( A \) be the event that a student is from XII A.
- Let \( B \) be the event that a student is from XII B.
- Let \( E \) be the event that a student has passed the examination.

### Step 5: Calculate the probabilities of events A and B
- Probability of A (P(A)) = Number of students in XII A / Total number of students
\[ P(A) = \frac{60}{100} = \frac{3}{5} \]
- Probability of B (P(B)) = Number of students in XII B / Total number of students
\[ P(B) = \frac{40}{100} = \frac{2}{5} \]

### Step 6: Calculate the conditional probabilities
- Probability of passing given A (P(E|A)) = Students passed in XII A / Total students in XII A
\[ P(E|A) = \frac{45}{60} = \frac{3}{4} \]
- Probability of passing given B (P(E|B)) = Students passed in XII B / Total students in XII B
\[ P(E|B) = \frac{20}{40} = \frac{1}{2} \]

### Step 7: Use Bayes' theorem to find the required probability
We need to find \( P(A|E) \), the probability that a student is from XII A given that they passed.
Using Bayes' theorem:
\[ P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B)} \]

### Step 8: Substitute the values into the formula
\[ P(A|E) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{3}{4}\right)}{\left(\frac{3}{5} \cdot \frac{3}{4}\right) + \left(\frac{2}{5} \cdot \frac{1}{2}\right)} \]

### Step 9: Calculate the numerator and denominator
- Numerator:
\[ = \frac{3}{5} \cdot \frac{3}{4} = \frac{9}{20} \]
- Denominator:
\[ = \frac{9}{20} + \frac{2}{10} = \frac{9}{20} + \frac{4}{20} = \frac{13}{20} \]

### Step 10: Final calculation
\[ P(A|E) = \frac{\frac{9}{20}}{\frac{13}{20}} = \frac{9}{13} \]

### Final Answer
Therefore, the probability that a student is from XII A given that they passed is:
\[ \frac{9}{13} \]