To solve this problem, we need to apply the concept of probability, specifically using the conditional probability approach. Here's how you can think through the problem step-by-step:

1. Understanding the Problem:
   • A bag contains 10 balls, where $k$ are red and $(10 - k)$ are black.
   • We are given that three balls drawn are all black, and we need to find the probability that the bag contains 1 red and 9 black balls.
2. Define the Events:
   • Let \(R_k\) be the event that the bag contains $k$ red balls.
   • Let \(B_3\) be the event that all three drawn balls are black.
3. Apply Conditional Probability:

We need to find \(P(R_1 \mid B_3)\), the probability that the bag contains 1 red ball given that all three drawn balls are black.

Using Bayes' theorem:

1. \(P(R_1 \mid B_3) = \frac{P(B_3 \mid R_1) \cdot P(R_1)}{P(B_3)}\)
2. Compute Probabilities:
   • Probability of drawing 3 black balls given 1 red ball: \(P(B_3 \mid R_1) = \frac{\binom{9}{3}}{\binom{10}{3}} = \frac{84}{120} = \frac{7}{10}\)
   • Probability of 1 red ball in the bag: \(P(R_1) = \frac{1}{11}\) (since each $k$ from 0 to 10 is equally likely).
   • Probability of drawing 3 black balls:

Calculate \(P(B_3)\) by summing over all possible values of $k$:

• \(P(B_3) = \sum_{k=0}^{10} P(B_3 \mid R_k) \cdot P(R_k)\) 
  where:
  • For any given $k$, \(P(B_3 \mid R_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}\)

1. Calculate Conditional Probability:

\(P(R_1 \mid B_3) = \frac{\frac{7}{10} \times \frac{1}{11}}{\frac{6}{11}} = \frac{7}{60} \times \frac{11}{6} = \frac{7}{55}\)

1. Conclusion:

Therefore, the probability that the bag contains 1 red and 9 black balls given that all three drawn balls are black is \(\frac{7}{55}\).