Q: Probability of a pair of eggs being one male and one female If I am given two fertilised eggs, what is the probability that they will hatch as a mixed pair?
(a) As a non-mathematician, I can see there are four permutations:
MM, FF, MF, FM and so the probability would be 1 in 2. 
Or, (b) if one egg hatches as, say, a female, there is a 1 in 2 chance the other will be male.
(c) However another argument is that (before any eggs hatch) there are three possible combinations:
MM, FF, MF and so there is a 1 in 3 chance of a mixed pair.
Which is correct?  Or is it that there is a 1 in 3 chance, but when one hatches, the probability changes to 1 in 2, much as for the Monty Hall problem?
Apologies if this is a nonsense question.
Thanks.

A: Answers a),b) are correct. 
a) Assuming that each egg is assigned M/F independently, label the eggs 1 and 2. Then the probability of MM,FF,MF,FM is 1/4 each. Therefore the total probability of having one M and one F is 1/2.
b) This argument works as you can pick egg 1, wait for it to hatch and then you know there is 1/2 chance the second egg will differ.
The reasoning for c) is incorrect. 
c) There are three possible combinations (if you don't label the eggs), but there are more ways for MF (/FM) to occur. It is not the case that every event is equally weighted. 

A: The question is about 'one male and one female'. It can be clearly seen there are two ways to do this. Since we treat two eggs as distinct ones, you have to consider all possible combinations, taking into account the eggs are distinct. So the third approach is wrong. Also, as mentioned in one of the comments, if you notice that in (c) the outcomes are not equally likely, then also you can reach the correct answer.

A: Your (b) is correct.
Your (c) incorrectly assumes that, because you've divided the possible outcomes into three categories, that each of these categories occurs with probability $\frac13$. There is no reason in this case for it to hold, but this assumption is often implicitly used in questions on probability: two sides to a (fair) coin, six sides to a (fair) die, an equal probability of a girl being born and a boy being born.
Your (a) is correct, although your reasoning in (c) demonstrates that you haven't yet quite understood why (a) is correct: in the case of births, the assumption that both outcomes are equally likely is warranted; and furthermore the genders of the two births are (presumably) independent, which means that also the four outcomes FF, FM, MF, MM are equally likely.
