To find the probability $P(\text{A and A})$, we need to consider the two events of drawing an "A" both times, with replacement.

Step 1: Total number of tiles in the bag

The total number of tiles in the bag is: $5 \, (\text{A's}) + 2 \, (\text{B's}) + 4 \, (\text{C's}) = 11 \, \text{tiles}$

Step 2: Probability of drawing an "A" on the first draw

The probability of drawing an "A" on the first draw is: $P(\text{A on 1st draw}) = \frac{5}{11}$

Step 3: Probability of drawing an "A" on the second draw

Since the tile is replaced after the first draw, the probability of drawing an "A" on the second draw is the same as the first: $P(\text{A on 2nd draw}) = \frac{5}{11}$

Step 4: Probability of drawing "A" on both draws (with replacement)

The probability of both events (drawing an "A" on the first and second draws) is the product of the individual probabilities: $P(\text{A and A}) = P(\text{A on 1st draw}) \times P(\text{A on 2nd draw}) = \frac{5}{11} \times \frac{5}{11} = \frac{25}{121}$

Final Answer:

The probability of drawing an "A" on both draws is: $P(\text{A and A}) = \frac{25}{121}$

Would you like a further explanation or have any questions on this? Here are some related topics you might find helpful:

1. What is the probability of drawing two "B"s?
2. How does replacement affect the probability in this type of problem?
3. What is the probability of drawing one "A" and one "C" in two draws?
4. How can you calculate the probability of drawing a "B" or "C" twice?
5. What would happen if the tiles were not replaced after each draw?

Tip: When dealing with replacement, the probability for each event remains unchanged because the total number of outcomes stays constant.