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A family has three children, all of whom are either a boy or a girl. Two of the kids are boys. What is the probability that the third child is a girl? Answer as a common fraction.

The problem is, it doesn't specify what chances of any gender being chosen, so I am not sure about it being 1/2.

What if they are identical boys?

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    Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. What do you think the probability that the third child is a girl is? Why? Please edit your question to include this information. – 2017-02-13
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    Also, it is usually a good idea to look around and see if a question like your question was asked before ... Which it has been .. Many times in fact ... On the right, under 'Related', you will probably find lots of other posts that should help you out. Good luck! – 2017-02-13

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The problem is, it doesn't specify what chances of any gender being chosen, so I am not sure about it being 1/2.

It is not actually justified by real life statistics, but lacking any such data to work with in these technique exercises, the student is expected to assume that each birth has an independent and unbiased probability of yielding a boy.   Ie: $1/2$ is okay.

As commenters have made clear, the question hinges entirely on whether we know which two are boys, and how we know this about them.

If we know two particular children are boys, and wish to know the probability that the third is a girl, then it is simply: $1/2$.

However, if we merely know that at least two of the children are boys, such as perhaps being told that two randomly selected children were boys, then we don't have any information to specify which two of the three children they are and so we don't know which child is the 'third' child.

Let $B$ represent the Count of Boys in the family of three.   Then $B$ is binomially distributed. $~\mathsf P(B{=}k)=\tbinom 3 k \tfrac 18$

$$\mathsf P(B{=} 2\mid B{\geq} 2) ~=~ \dfrac{\mathsf P(B{=}2)}{\mathsf P(B{=}2)+\mathsf P(B{=}3)} ~=~ \dfrac{\frac 38}{\frac 38 +\frac 18} ~=~ \dfrac{3}{4}$$

Which can be visualised by considering the equally likely birth orders: $$\cal \{ \color{blue}{BBB}, \color{blue}{BB}\color{brown}G, \color{blue}B\color{brown}G\color{blue}B, \color{brown}G\color{blue}{BB} , \ldots \}$$

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Classic problem...

The answer is 1/2. The fact that the first two children were boys has nothing to do with the third child's gender.

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    Presumably the question is intended to be interpreted as "*given that at least two of the children are boys, what is the probability that all three of the children are boys*" in which case the probability is not $\frac{1}{2}$ as opposed to the wholly uninteresting problem "*given that the youngest two children are boys, what is the probability that the oldest child is also a boy*" – 2017-02-13
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    I'm taking it as two specified children being boys and asking for the last one. Otherwise it's 3/4 because we have GBB/BGB/BBG/BBB with 3/4 chance of having a girl. – 2017-02-13
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    You could argue that the wording given is in fact saying "*exactly two of the three children are boys, what is the probability that all three are boys?*" in which case the probability is in fact zero. Wording is incredibly important to this problem, and the only interesting problem is the first one I mention since it gives a counterintuitive answer. The problem just says "two of the kids are boys", it didn't specify **which** two. – 2017-02-13
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    In that case, it's the problem itself... so pretty much depending on what's being asked the answer is just about anything you can think of. – 2017-02-13
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    What if the boys are identical? – 2017-02-13