Two balls are drawn in succession out of a box containing 3 red and 4 white balls. Find the probability that at least 1 ball was red, given that the first ball was (A) Replaced before the second draw. (B) Not replaced before the second draw. (A) Find the probability that at least 1 ball was red, given that the first ball was replaced before the second draw. $\frac{33}{49}$ (Simplify your answer. Type an integer or a fraction.) (B) Find the probability that at least 1 ball was red, given that the first ball was not replaced before the second draw. (Simplify your answer. Type an integer or a fraction.)

Step 1 ：For part (A), since the first ball is replaced before the second draw, the two events are independent. The probability that at least one ball is red is the same as 1 minus the probability that both balls are white. There are 3 red balls and 4 white balls, so the total number of balls is 7. The probability of drawing a white ball is \(\frac{4}{7}\). Since the balls are replaced, the probability of drawing a white ball in the second draw is also \(\frac{4}{7}\). So the probability of drawing two white balls is \(\frac{4}{7} * \frac{4}{7} = \frac{16}{49}\). Therefore, the probability of drawing at least one red ball is \(1 - \frac{16}{49} = \frac{33}{49}\).

Step 2 ：For part (B), since the first ball is not replaced before the second draw, the two events are dependent. If the first ball is red, there are 2 red balls and 4 white balls left, so the probability of drawing a white ball in the second draw is \(\frac{4}{6} = \frac{2}{3}\). If the first ball is white, there are 3 red balls and 3 white balls left, so the probability of drawing a red ball in the second draw is \(\frac{3}{6} = \frac{1}{2}\). Therefore, the probability of drawing at least one red ball is \(1 - \frac{2}{3} * \frac{1}{2} = \frac{5}{6}\).

Step 3 ：The probability that at least 1 ball was red, given that the first ball was replaced before the second draw is \(\boxed{\frac{33}{49}}\).

Step 4 ：The probability that at least 1 ball was red, given that the first ball was not replaced before the second draw is \(\boxed{\frac{5}{6}}\).