The following table shows the blood groups of 60 students of a class :
`|{:("Blood groups A B O AB"),("Number of students 16 12 23 9"):}|`
One student of the class is chosen at random. What is the probability that the chosen student has blood group?
`"(i) O (ii) AB (iii) A (iv) B"`

Answer

To solve the problem, we need to calculate the probability of selecting a student with each specified blood group from the given data. The probability \( P \) of an event is calculated using the formula:

\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]

### Step-by-Step Solution:

1. **Identify the total number of students**:
The total number of students in the class is given as 60.

2. **Calculate the probability for blood group O**:
- Number of students with blood group O = 23
- Total students = 60
- Probability \( P(O) \) = \( \frac{23}{60} \)

3. **Calculate the probability for blood group AB**:
- Number of students with blood group AB = 9
- Total students = 60
- Probability \( P(AB) \) = \( \frac{9}{60} \)
- Simplifying \( \frac{9}{60} \) gives \( \frac{3}{20} \)

4. **Calculate the probability for blood group A**:
- Number of students with blood group A = 16
- Total students = 60
- Probability \( P(A) \) = \( \frac{16}{60} \)
- Simplifying \( \frac{16}{60} \) gives \( \frac{4}{15} \)

5. **Calculate the probability for blood group B**:
- Number of students with blood group B = 12
- Total students = 60
- Probability \( P(B) \) = \( \frac{12}{60} \)
- Simplifying \( \frac{12}{60} \) gives \( \frac{1}{5} \)

### Final Answers:
- (i) Probability of blood group O: \( \frac{23}{60} \)
- (ii) Probability of blood group AB: \( \frac{3}{20} \)
- (iii) Probability of blood group A: \( \frac{4}{15} \)
- (iv) Probability of blood group B: \( \frac{1}{5} \)