To solve this problem, we calculate the probability of selecting a sample of five balls that includes both the $3$-ball and the $8$-ball, out of a total of 10 balls, when drawn without replacement.

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Step 1: Total number of ways to select 5 balls

The total number of ways to choose 5 balls from 10 is given by the combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ Here, $n = 10$ and $k = 5$: $\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252$ Thus, there are 252 ways to select 5 balls from 10.

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Step 2: Ways to include the $3$-ball and $8$-ball

To include both the $3$-ball and the $8$-ball in the selection:

1. The $3$-ball and $8$-ball are fixed, leaving 3 more balls to be chosen from the remaining 8 balls ($10 - 2 = 8$).
2. The number of ways to choose 3 balls from 8 is: $\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$

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Step 3: Probability of including the $3$-ball and $8$-ball

The probability is the ratio of favorable outcomes to total outcomes: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{\binom{8}{3}}{\binom{10}{5}}$ Substituting the values: $P = \frac{56}{252} = \frac{2}{9}$

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Final Answer:

The probability that the $3$-ball and $8$-ball are in the sample is: $\boxed{\frac{2}{9}}$

Let me know if you want further details or clarifications. 😊

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Related Questions:

1. How do combinations differ from permutations in such problems?
2. What is the probability if balls are drawn with replacement?
3. How would the probability change if we wanted either the $3$-ball or $8$-ball (not both)?
4. What is the probability of selecting exactly 4 specific balls in a sample of 5?
5. How does the probability change if the total number of balls is increased to 12?

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Tip:

When solving probability problems, ensure to distinguish between events occurring independently (with replacement) versus dependently (without replacement) to set up the correct formula.