The problem in the image is in Indonesian and it translates as follows:

Problem 1: It is known that 10 pieces of paper are numbered from 1 to 10. If 2 pieces of paper are taken and the numbers on these pieces of paper are summed, what is the probability that the sum of the numbers will be odd?

To solve this:

1. Odd and even number properties:

   • The sum of two numbers is odd if and only if one of them is odd and the other is even.
   • The sum of two odd or two even numbers is always even.

2. Classifying the numbers 1 to 10:

   • Odd numbers: {1, 3, 5, 7, 9} (There are 5 odd numbers.)
   • Even numbers: {2, 4, 6, 8, 10} (There are 5 even numbers.)

3. Counting favorable outcomes:

   • To get an odd sum, we need one odd and one even number.
   • The number of ways to choose 1 odd number from 5 odd numbers is $\binom{5}{1} = 5$.
   • The number of ways to choose 1 even number from 5 even numbers is $\binom{5}{1} = 5$.
   • So, the total number of favorable outcomes (odd + even) is $5 \times 5 = 25$.

4. Total number of possible outcomes:

   • The total number of ways to choose any 2 numbers from 10 is $\binom{10}{2} = 45$.

5. Probability:

   • The probability of getting an odd sum is the ratio of favorable outcomes to total outcomes: $P(\text{odd sum}) = \frac{25}{45} = \frac{5}{9}$

Therefore, the probability of getting a sum that is odd is $\frac{5}{9}$.

Would you like any further details or clarifications?

Here are 5 related questions to deepen understanding:

1. What would the probability be if we added another number to the set?
2. How would the result change if instead of 2 numbers, 3 numbers were selected?
3. What if only even numbers were allowed in the selection—what would the new probability be?
4. Can you calculate the probability of getting a sum that is divisible by 3?
5. How would you approach this problem if it involved 15 pieces of paper numbered from 1 to 15?

Tip: When dealing with probabilities, always break down the problem by analyzing how the event (in this case, the sum) behaves under specific conditions (like even and odd numbers).