A fair coin is tossed three times. What is the probability that all three tosses land on heads, given that:
(a) the first toss lands on heads
(b) at least one toss lands on heads
For part a) I think the prob of A given B = $\frac{P(A∩B)}{P(B)}$ and I know P(A) = $1/8$ and P(B) = $1/2$ so would a) be $\frac{\left(\frac{1}{8}\cdot \:\frac{1}{2}\right)}{\frac{1}{2}}$ which is $\frac{1}{8}$
for part b) I don't really understand how this question is differing from a)
For the first, your sample space is $Hxx$ where the $X$s represent either heads or tails. How many events are there? How many are favorable?
For the second, your sample space is all three toss events except $TTT$ because in all the other cases there is at least one head. How many events? How many are favorable.
In your calculation, $A$ guarantees $B$, so $P(A \cap B)=P(A)=\frac 18$.