Let's break this down step by step.

Total Number of Ways to Draw 3 Balls

The box contains 11 balls numbered from 1 to 11. When drawing 3 balls out of 11, we need to calculate the total number of combinations.

The number of ways to choose 3 balls from 11 is given by the combination formula:

$\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$

So, the total number of ways to draw 3 balls is 165.

When is the Sum Odd?

The sum of the numbers on the balls will be odd if the number of odd balls drawn is odd. This is because:

• The sum of an odd number of odd numbers is odd (e.g., 1 + 3 + 5 = 9).
• The sum of an even number of odd numbers is even (e.g., 1 + 3 + 2 = 6).

Counting Odd and Even Balls

• The odd-numbered balls are: 1, 3, 5, 7, 9, 11. So, there are 6 odd balls.
• The even-numbered balls are: 2, 4, 6, 8, 10. So, there are 5 even balls.

For the sum to be odd, we must draw 1 or 3 odd balls (because the sum of 1 odd number is odd, and the sum of 3 odd numbers is odd).

Case 1: Drawing 1 Odd and 2 Even Balls

• Number of ways to choose 1 odd ball from 6 odd balls: $\binom{6}{1} = 6$
• Number of ways to choose 2 even balls from 5 even balls: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$

So, the total number of ways to draw 1 odd ball and 2 even balls is:

$6 \times 10 = 60$

Case 2: Drawing 3 Odd Balls

• Number of ways to choose 3 odd balls from 6 odd balls: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

So, the total number of ways to draw 3 odd balls is 20.

Total Number of Favorable Outcomes

The total number of favorable outcomes (the sum being odd) is the sum of the results from Case 1 and Case 2:

$60 + 20 = 80$

Probability Calculation

Now, the probability of drawing 3 balls such that the sum of the numbers is odd is the ratio of favorable outcomes to total outcomes:

$P(\text{odd sum}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{80}{165}$

Simplifying the fraction:

$\frac{80}{165} = \frac{16}{33}$

Thus, the probability that the sum of the numbers on the balls drawn is odd is $\frac{16}{33}$.

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Would you like further details on any part of the solution, or do you have other questions?

Related Questions:

1. What is the probability of drawing 3 balls such that the sum is even?
2. How do you calculate the number of combinations when drawing more than 3 balls?
3. What happens if the total number of balls is changed to something other than 11?
4. How can you apply this kind of probability problem to other real-life scenarios?
5. Can this method be used to calculate probabilities for more complex sets (like with multiple categories)?

Tip:

For problems like these, always start by identifying the conditions that make the outcome desirable (odd sum) and then break the problem down into manageable cases (1 odd + 2 even, or 3 odd).