a) Let A be the event that there is at least one head in the three tosses, and B be the event that there are at least two heads.
We want to compute P(B|A), the probability of B given A. By the definition of conditional probability, we have P(B|A) = P(A∩B) / P(A).
First, let's compute the probabilities of the complementary events A' and B', which are "no heads in the three tosses" and "fewer than two heads", respectively.
For A': the only way to get no heads is to have the coin land tails three times. This happens with probability (1/3)^3 = 1/27. Since P(A) + P(A') = 1, we get P(A) = 26/27.
For B': there can be either no heads or exactly one head. We already computed the probability of no heads, which is 1/27. To get exactly one head, there are three scenarios: H-T-T, T-H-T, or T-T-H. The probability of each of these scenarios is (2/3)*(1/3)*(1/3) = 2/27. In total, P(B') = P(no heads) + P(exactly one head) = 1/27 + 3*(2/27) = 1/3. Since P(B) + P(B') = 1, we get P(B) = 2/3.
Now we can compute the intersection of A and B, A∩B. We know that B implies at least two heads, so A∩B is just the event "at least two heads". We know that P(at least two heads) = P(B) = 2/3.
Finally, we can compute the conditional probability P(B|A) = P(A∩B) / P(A) = (2/3) / (26/27) = (2/3)*(27/26) = 9/13.
b) Let C be the event of exactly one head. To find P(C|A), notice that C is the complement of B given A, since if there is at least one head and there are not at least two heads, then there must be exactly one head. Thus, P(C|A) = 1 - P(B|A) = 1 - 9/13 = 4/13.