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The problem describes a survey conducted in a class where:

- 68 students eat breakfast with rice,
- 50 students eat breakfast with bread,
- 8 students eat breakfast with both rice and bread,
- 35 students do not eat breakfast with rice or bread.

We are asked to determine the probability that a randomly selected student from the class only eats breakfast with rice.

### Let's break it down:
Let:
- \( A \) represent the set of students who eat breakfast with rice.
- \( B \) represent the set of students who eat breakfast with bread.
- The number of students who eat both rice and bread is 8.
- The number of students who do not eat either rice or bread is 35.

The goal is to find the number of students who only eat rice and use that to calculate the probability.

### Step 1: Calculate the total number of students in the class.
We can calculate the total number of students as the sum of all groups:
\[
\text{Total students} = (\text{students who eat rice}) + (\text{students who eat bread}) + (\text{students who eat neither}) - (\text{students who eat both})
\]
This is because the students who eat both rice and bread have been counted twice.

\[
\text{Total students} = 68 + 50 - 8 + 35 = 145
\]

### Step 2: Calculate the number of students who only eat rice.
To find the number of students who only eat rice, subtract those who eat both rice and bread from the total number of students who eat rice:
\[
\text{Only rice} = 68 - 8 = 60
\]

### Step 3: Calculate the probability.
The probability of selecting a student who only eats rice is:
\[
P(\text{only rice}) = \frac{\text{Number of students who only eat rice}}{\text{Total number of students}} = \frac{60}{145}
\]

Thus, the probability that a randomly selected student only eats rice is \( \frac{60}{145} \).

Would you like further details or clarifications on this problem?

Here are five related questions you might find interesting:
1. How would the probability change if the number of students who eat both rice and bread increased to 12?
2. What is the probability that a randomly selected student does not eat rice or bread?
3. How can we represent this problem using a Venn diagram?
4. If a student is known to eat bread, what is the probability that they also eat rice?
5. How would the probability change if the total number of students in the class were different?

**Tip:** When solving problems involving sets and probabilities, Venn diagrams can be very useful for visualizing the relationships between different groups.

Dalam suatu kelas dilakukan survey kepada para siswa. Dari survey ini diperoleh data 68 orang sarapan dengan nasi, 50 orang sarapan dengan roti, dan 8 orang sarapan nasi dan roti, sedangkan 35 orang sarapan tidak dengan nasi atau roti. Jika dalam kelas tersebut dipilih seorang anak, peluang yang terpilih adalah anak yang hanya sarapan dengan nasi adalah ....

This problem involves calculating the probability that a randomly selected student from a class of 145 students only eats rice for breakfast. It covers fundamental probability concepts and set theory, suitable for students in Grades 7-9.

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