State for each of the following relations whether it is function or not. (Write Yes or No)
`{(x,y):x inB,yinP,y" is a passenger on x"}` where B is the set of buses of a school and P is set of pupils of some schools.

Answer

To determine whether the given relation is a function, we need to analyze the definition of a function and the specifics of the relation provided.

### Step-by-Step Solution:

1. **Understanding the Sets**:
- Let \( B \) be the set of buses: \( B = \{ B_1, B_2, B_3, B_4 \} \).
- Let \( P \) be the set of pupils: \( P = \{ P_1, P_2, P_3, P_4 \} \).

2. **Defining the Relation**:
- The relation is defined as \( R = \{ (x, y) : x \in B, y \in P, \text{ and } y \text{ is a passenger on } x \} \).
- This means that for each bus \( x \), there can be one or more pupils \( y \) who are passengers on that bus.

3. **Checking the Function Criteria**:
- A relation is a function if every element in the domain (in this case, the set of buses \( B \)) is associated with exactly one element in the codomain (the set of pupils \( P \)).
- In simpler terms, for each bus \( x \), there should be only one pupil \( y \) associated with it.

4. **Analyzing the Relation**:
- Consider the possibility that multiple pupils can be passengers on the same bus. For example, pupil \( P_1 \) can be on bus \( B_1 \) and pupil \( P_2 \) can also be on bus \( B_1 \).
- This indicates that a single bus \( B_1 \) can have multiple pupils associated with it.

5. **Conclusion**:
- Since one bus can have multiple passengers, this violates the definition of a function where each input (bus) must map to exactly one output (pupil).
- Therefore, the relation \( R \) is **not a function**.

### Final Answer:
**No**, the relation is not a function.