A fair coin is tossed 12 times.
(A) What is the probability of tossing a head on the 12th toss, given that the preceding 11 tosses were tails?
(B) What is the probability of getting either 12 heads or 12 tails?
(A) What is the probability of tossing a head on the 12th toss, given that the preceding 11 tosses were tails?
(Type an integer or a fraction. Simplify your answer.)
(B) What is the probability of getting either 12 heads or 12 tails?
(Type an integer or a fraction. Simplify your answer.)

Step 1 ：The problem is asking for two probabilities related to tossing a fair coin 12 times.

Step 2 ：For part (A), the probability of tossing a head on the 12th toss, given that the preceding 11 tosses were tails, is independent of the previous tosses. This is because each coin toss is an independent event. Therefore, the probability is simply the probability of tossing a head, which is \(\frac{1}{2}\).

Step 3 ：For part (B), the probability of getting either 12 heads or 12 tails is the sum of the probabilities of getting 12 heads and getting 12 tails. Since each coin toss is an independent event, the probability of getting 12 heads is \((\frac{1}{2})^{12}\) and the probability of getting 12 tails is also \((\frac{1}{2})^{12}\). Therefore, the total probability is \(2*(\frac{1}{2})^{12}\).

Step 4 ：Final Answer: For part (A), the probability of tossing a head on the 12th toss, given that the preceding 11 tosses were tails, is \(\boxed{\frac{1}{2}}\).

Step 5 ：For part (B), the probability of getting either 12 heads or 12 tails is \(\boxed{\frac{1}{2048}}\).