Final answer:
In problem a), two events are independent, and the probability of getting an A first and an A second is 4/49. In problem b), the events are not independent, and the probability of getting an A first and an A second is 1/21.
Step-by-step explanation:
For problem a), since Paola replaces the tile after drawing, the two events are independent. The probability of drawing an A on the first draw is 2 out of 7 (2/7), because there are 2 As in a bag of 7 tiles. Since she replaces the tile, the probability of drawing an A on the second draw remains the same at 2/7. Thus, to find the probability of getting an A first and then an A second, we multiply the probabilities of each event (2/7 * 2/7), which equals 4/49.
For problem b), when Paola does not replace the tile after the first draw, the events are not independent. The probability of drawing an A on the first draw is still 2/7, but if she draws an A and does not replace it, there is now 1 A out of 6 remaining tiles. Thus, the probability of drawing an A second is now 1/6. To find the probability of getting an A first and then an A second, we multiply the probabilities of the two events (2/7 * 1/6), which results in 2/42 or 1/21.