A fair coin is flipped 3 times. Probability of all $3$ heads given at least $2$ were heads Here's a problem from Prof. Blitzstein's Stat 110 textbook. Please see my solution below.

A fair coin is flipped 3 times. The toss results are recorded on separate slips of paper (writing “H” if Heads and “T” if Tails), and the 3 slips of paper are thrown into a hat.
(a) Find the probability that all 3 tosses landed Heads, given that at least 2 were Heads.
(b) Two of the slips of paper are randomly drawn from the hat, and both show the letter H. Given this information, what is the probability that all 3 tosses landed Heads?

The binomial distribution of the number of heads (#H) is as follows:
$\mathbb{P}(\#H=0) = 1/8 $
$\mathbb{P}(\#H=1) = 3/8 $
$\mathbb{P}(\#H=2) = 3/8 $
$\mathbb{P}(\#H=3) = 1/8 $
a) Probability of 3 heads given at least 2 heads = $\mathbb{P}(\#H=3)/\mathbb{P}(\#H \ge 2 ) = (1/8)/(1/2)=1/4$
b) But I'm not able to understand how (b) is different from (a). To me it's the same problem and has the same answer.
Can anyone help please?

Perhaps it's worth emphasising that the difference is that in (a) the condition is that any $2$ coins landed heads and in (b) the condition is that $2$ particular coins landed heads. Condition (b) is more specific; it leaves only $2$ possible configurations, whereas (a) leaves $4$. Since in each case you're looking for the probability of one of these configurations occurring, the probabilities are $\frac12$ and $\frac14$.
