To calculate $P(A \text{ and } A)$, we follow these steps:

Step 1: Define the situation

• The probability of selecting an $A$ from the bag is the ratio of the number of $A$'s to the total number of tiles: $P(A) = \frac{\text{Number of A's}}{\text{Total number of tiles}} = \frac{5}{5+4+4} = \frac{5}{13}.$

• Since the tile is replaced after the first draw, the probabilities remain the same for the second draw. Therefore, $P(A)$ is still $\frac{5}{13}$ for the second draw.

Step 2: Use the multiplication rule for independent events

The probability of both events (drawing an $A$ on the first and second draws) is: $P(A \text{ and } A) = P(A) \times P(A) = \frac{5}{13} \times \frac{5}{13} = \frac{25}{169}.$

Final Answer:

$P(A \text{ and } A) = \frac{25}{169}.$

Would you like me to explain further or explore variations of this problem?

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Related Questions:

1. What is $P(A \text{ and } B)$ in the same scenario?
2. What is the probability of drawing two different tiles, $P(A \text{ and not } A)$?
3. If the first tile is not replaced, how does $P(A \text{ and } A)$ change?
4. What is the probability of selecting a $B$ on both draws?
5. How can the total probabilities for all outcomes confirm the law of total probability?

Tip:

When events are independent, the probabilities of sequential events can be multiplied directly.