To solve this problem, let's break it down step by step.

Step 1: Understand the situation

• The bag contains:
  • $5$ A's,
  • $4$ B's,
  • $2$ C's.
• Total number of tiles = $5 + 4 + 2 = 11$.

Step 2: Define the probabilities

• Probability of drawing a $C$ (on any draw): $P(C) = \frac{\text{Number of C tiles}}{\text{Total number of tiles}} = \frac{2}{11}.$

Step 3: Multiply probabilities (since the draws are independent)

Since the first tile is replaced, the probability of drawing $C$ on the second draw is the same as the first draw. Thus: $P(\text{C and C}) = P(C) \cdot P(C) = \frac{2}{11} \cdot \frac{2}{11} = \frac{4}{121}.$

Final Answer:

$P(\text{C and C}) = \frac{4}{121}.$

Would you like a detailed explanation of the independence concept or have other questions? 😊

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Here are 5 related questions for practice:

1. What is $P(\text{Upper A and Upper C})$ in the same scenario?
2. How does the probability change if the first tile is not replaced?
3. What is the probability of getting $B$ on the first draw and $A$ on the second draw?
4. What is the probability of not drawing $C$ on either of the two draws?
5. What is the total probability of drawing either $A$ or $B$ on the first draw?

Tip: For probability problems, always double-check whether the events are independent or dependent!