There are in total of 40 students. 14 are taking Maths and 29 are taking computer. Therefore there have to be 3 students who are taking both the classes. So, 29 – 3 = 26 students are taking only Computer. So probability = 26/40 = 13/20 = 65%

It looks like you're trying to calculate the probability of students taking only Computer classes based on the given numbers. However, let's clarify the situation step by step. 1. **Total Students**: 40 2. **Students taking Maths**: 14 3. **Students taking Computer**: 29 4. **Students taking both Maths and Computer**: 3 From this information, we can determine the following: - **Students taking only Computer**: \[ \text{Students taking only Computer} = \text{Total Computer Students} - \text{Students taking both} = 29 - 3 = 26 \] - **Students taking only Maths**: \[ \text{Students taking only Maths} = \text{Total Maths Students} - \text{Students taking both} = 14 - 3 = 11 \] - **Students taking neither subject**: \[ \text{Students taking neither} = \text{Total Students} - (\text{Students taking only Computer} + \text{Students taking only Maths} + \text{Students taking both}) = 40 - (26 + 11 + 3) = 40 - 40 = 0 \] Now, the probability calculations you've performed seem to be for the scenario of choosing a student who is taking only Computer classes. - The probability that a randomly selected student is taking only Computer classes is: \[ P(\text{Only Computer}) = \frac{\text{Students taking only Computer}}{\text{Total Students}} = \frac{26}{40} = \frac{13}{20} \] This is indeed 65%. So, your calculations are correct for finding the probability of selecting a student who takes only Computer classes: **65% (or \( \frac{13}{20} \))**.