The fact that the first child among two is a boy has no effect on the sex of the second because the draw "sex of the first child" and "sex of the second child" are independant.
The probability of the other sibling being a girl is equal to the probability for any child picked at random to be a girl.
The chance is 1/2.
Another form of the question would be:
You play heads or tails -with a balanced coin, no loophole- twice.
The first time you play you get heads, what is the chance for the
second time you play for it to get tails ?
It is 1/2. The chance for you to get tails is not increased by the
fact that you had heads previously.
As soon as we know that the first child is a male, FM and FF become impossible, leaving MF and MM.
EDIT: I see that I still am not the accepted answer. And after discussing with my friend I can see why.
My answer is counter intuitive because in real life people tend to apply the Law of Large Numbers -or law of averages- to their everyday / low amount of draws routines. For instance:
You are at the casino, and play roulette.
You place your bets on red and lose 3 times.
You apply the law of average and believe that now your chance to win
by betting red is very high. But this reasoning is irrational. The
roulette has no memory of the previous rolls. Your chances by betting
red or black are still the same.
Let me make another attempt at making you see that Skywalker's answer is flawed.
A group of 4 male dudes come to you at a party and tell you that they have a 5th friend of which they ask you to guess the gender.
According to the logic used in his argument the chance for the 5th friend to be a girl is 4/5 = 0.80 = 80% because the possible combinations are
MMMMM MMMMF MMMFM MMFMM MFMMM FMMMM
The fact that the 4 other dudes are males bring no knowledge of the gender of the 5th.
The chance in that case is 1/2 too.