In the class of 30 children, there are 20 who can swim and 25 who can ride a bike. There are no chil who cannot swim nor ride a bike.

To solve the problem, we will analyze the information given about the class of 30 children using a Venn diagram, and then calculate the probabilities required.

1. **Understanding the Information**:
- Total children: 30
- Children who can swim: 20
- Children who can ride a bike: 25
- No children can be described as neither swimming nor riding a bike.

2. **Setting Up the Venn Diagram**:
We can denote:
- \( S \): the set of children who can swim
- \( B \): the set of children who can ride a bike

To find the overlap (children who can both swim and ride a bike), we need to apply the principle of inclusion-exclusion:

\[
|S \cup B| = |S| + |B| - |S \cap B|
\]

Here, \( |S \cup B| \) is the total number of unique children (which is 30).

Plugging in the numbers we have:
\[
30 = 20 + 25 - |S \cap B|
\]
\[
30 = 45 - |S \cap B|
\]
\[
|S \cap B| = 45 - 30 = 15
\]

Thus, there are 15 children who can both swim and ride a bike.

3. **Finding the Exclusive Sets**:
- Children who can only swim (but not ride a bike):
\[
|S| - |S \cap B| = 20 - 15 = 5
\]
- Children who can only ride a bike (but not swim):
\[
|B| - |S \cap B| = 25 - 15 = 10
\]

4. **Completing the Venn Diagram**:
- Children who can swim only: 5
- Children who can ride a bike only: 10
- Children who can do both: 15

To summarize:
- Only swim: 5
- Only bike: 10
- Both: 15

5. **Calculating Probabilities**:
Now, let’s compute the probabilities for each part:

(a) **Probability that a randomly selected child can swim and ride a bike**:
\[
P(\text{swim and bike}) = \frac{|S \cap B|}{\text{Total children}} = \frac{15}{30} = \frac{1}{2}
\]

(b) **Probability that a randomly selected child cannot swim but can ride a bike**:
- Since all children can swim or ride a bike, there is no child who cannot swim but can ride a bike.
\[
P(\text{not swim but bike}) = 0
\]

(c) **Probability that a randomly selected child can swim but cannot ride a bike**:
\[
P(\text{swim but not bike}) = \frac{|S| - |S \cap B|}{\text{Total children}} = \frac{5}{30} = \frac{1}{6}
\]

### Summary of Results:
- (a) Probability that a child can swim and ride a bike: \( \frac{1}{2} \)
- (b) Probability that a child cannot swim but can ride a bike: \( 0 \)
- (c) Probability that a child can swim but cannot ride a bike: \( \frac{1}{6} \)