We are given an urn with the following contents:

• 9 red balls,
• 5 white balls,
• 10 blue balls.

The total number of balls in the urn is: $9 + 5 + 10 = 24 \text{ balls}.$

We are asked to find the probability that at least one ball of each color is chosen when four balls are drawn randomly without replacement.

Step 1: Total number of ways to choose 4 balls

The total number of ways to choose 4 balls from the 24 available balls is given by the combination formula: $\binom{24}{4} = \frac{24 \times 23 \times 22 \times 21}{4 \times 3 \times 2 \times 1} = 10626.$

Step 2: Calculating the number of favorable outcomes

We need to find the number of favorable outcomes where at least one ball of each color is chosen. This means selecting:

• 1 red ball,
• 1 white ball,
• 1 blue ball,
• 1 additional ball (which could be of any color).

Substep 2.1: Case breakdown

We will break the problem into cases based on how the extra ball is chosen (either red, white, or blue).

• Case 1: One extra red ball.

  • Choose 2 red balls from 9: $\binom{9}{2}$,
  • Choose 1 white ball from 5: $\binom{5}{1}$,
  • Choose 1 blue ball from 10: $\binom{10}{1}$.

  The number of ways to choose in this case is: $\binom{9}{2} \times \binom{5}{1} \times \binom{10}{1} = \frac{9 \times 8}{2 \times 1} \times 5 \times 10 = 36 \times 5 \times 10 = 1800.$

• Case 2: One extra white ball.

  • Choose 1 red ball from 9: $\binom{9}{1}$,
  • Choose 2 white balls from 5: $\binom{5}{2}$,
  • Choose 1 blue ball from 10: $\binom{10}{1}$.

  The number of ways to choose in this case is: $\binom{9}{1} \times \binom{5}{2} \times \binom{10}{1} = 9 \times \frac{5 \times 4}{2 \times 1} \times 10 = 9 \times 10 \times 10 = 900.$

• Case 3: One extra blue ball.

  • Choose 1 red ball from 9: $\binom{9}{1}$,
  • Choose 1 white ball from 5: $\binom{5}{1}$,
  • Choose 2 blue balls from 10: $\binom{10}{2}$.

  The number of ways to choose in this case is: $\binom{9}{1} \times \binom{5}{1} \times \binom{10}{2} = 9 \times 5 \times \frac{10 \times 9}{2 \times 1} = 9 \times 5 \times 45 = 2025.$

Substep 2.2: Total favorable outcomes

Now, we add up all the favorable cases: $1800 + 900 + 2025 = 4725.$

Step 3: Probability

The probability is the ratio of favorable outcomes to total outcomes: $\text{Probability} = \frac{4725}{10626}.$

This simplifies (after dividing both numerator and denominator by their greatest common divisor, which is 3): $\text{Probability} = \frac{1575}{3542}.$

Thus, the probability that at least one ball of each color is chosen is: $\boxed{\frac{1575}{3542}}.$

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Do you need any further explanation or details on this problem? Here are some related questions for you to explore:

1. How would the probability change if the number of balls of each color in the urn were different?
2. How can we calculate the probability of drawing no balls of a certain color?
3. How would the problem change if we were drawing more or fewer than 4 balls?
4. How do combinations work in a situation like this with drawing multiple balls without replacement?
5. What is the difference between this type of problem and one involving replacement?

Tip: When working with probabilities, it’s always helpful to break down the problem into smaller, manageable cases and then combine the results.