Four fair coins are tossed one after another, and you are told, truthfully, that the FIRST coin came up heads. Given only this, what is the probability that exactly three of the four coins came up heads?

The report settles the first coin, so only the other three are still in doubt — $2^3=8$ equally likely outcomes. Three heads in all now means exactly two heads among those three, which happens in $C(3,2)=3$ ways. Probability = 3/8. Trap $1/4=C(4,3)/2^4=4/16$ answers the unconditional question and throws away the information given. Trap $1/8=(1/2{)}^3$ prices only one particular pattern such as H H H T and forgets that the odd tail can sit in any of three places. Trap 1/2 guesses without using either the condition or the count.