The problem asks for the probability that the sum of five randomly chosen cards from a set of nine (numbered 1 to 9) is an odd number.

Key information:

• There are 9 cards, numbered 1 to 9.
• A person picks 5 cards at random.
• The probability that the sum of the chosen 5 cards is odd is required.

Step-by-step solution:

1. Odd and even numbers among the cards:

   • Odd numbers: $1, 3, 5, 7, 9$ (5 cards in total)
   • Even numbers: $2, 4, 6, 8$ (4 cards in total)

2. Sum condition:

   • For the sum of the 5 cards to be odd, the number of odd cards picked must also be odd. This is because:
     • Odd number of odd cards results in an odd sum.
     • Even number of odd cards results in an even sum.

3. Possible cases: To have an odd sum, the number of odd cards in the selection must be 3 or 5. Let's calculate the number of ways to select cards under these conditions.

   • Case 1: 3 odd cards and 2 even cards:

     • Number of ways to choose 3 odd cards from 5: $\binom{5}{3} = 10$
     • Number of ways to choose 2 even cards from 4: $\binom{4}{2} = 6$
     • Total ways for this case: $10 \times 6 = 60$

   • Case 2: 5 odd cards and 0 even cards:

     • Number of ways to choose 5 odd cards from 5: $\binom{5}{5} = 1$
     • Number of ways to choose 0 even cards from 4: $\binom{4}{0} = 1$
     • Total ways for this case: $1 \times 1 = 1$

4. Total favorable outcomes: The total number of favorable outcomes (where the sum is odd) is: $60 + 1 = 61$

5. Total possible outcomes: The total number of ways to choose 5 cards from 9 is: $\binom{9}{5} = 126$

6. Probability: The probability of picking 5 cards such that their sum is odd is the ratio of favorable outcomes to the total outcomes: $P(\text{odd sum}) = \frac{61}{126}$

This is the probability that the sum of the five chosen cards is odd.

Would you like further details or clarifications? Here are 5 related questions:

1. How do we compute binomial coefficients like $\binom{5}{3}$?
2. Why does an odd number of odd numbers always lead to an odd sum?
3. How would the solution change if we needed the sum to be even instead of odd?
4. Can this concept be extended to larger sets of numbers?
5. What is the total number of combinations when choosing 4 cards instead of 5?

Tip: Binomial coefficients $\binom{n}{k}$ are used to calculate how many ways we can choose $k$ items from $n$ items without regard to the order.