tomwilliam said:
Thanks to everyone for helping!
Here's a idea/suggestion. Usually a question like this asks you to calculate only one thing. In this case, the probability of picking both twins. This might lead to only a partial understanding of these problems. When I'm learning I like to be more thorough in terms of (within reason) looking at the whole problem. And thereby seeing the whole picture. And, why not? If the aim is to learn probability theory, why just do problems in little bits and pieces?
In this case, I would have tried to see the whole picture. Here are my thought processes.
There are four outcomes:
1) You have both twins.
2) You have only the girl twin (and two others).
3) You have only the boy twin (and two others).
4) You have neither twin.
Observation: these are
mutually exclusive; and
represent all the possible outcomes. So, these four probabilities must add up to ##1##.
Observation: they are
not all equally likely;
Intuitive insight: 2) and 3) must have an equal probability. In more complicated problems, I would think carefully about relying on these insights. So, I might calculate 2) and 3) in any case and check they are the same. But, in this case, it seems clear to me that 2) and 3) must be equal.
My next thought is what is the easiest to calculate? It looks like 4). That must be:
$$p_4 = \frac {28} {30} \times \frac {27} {29} \times \frac {26} {28} = 0.807$$And that seems like a sensible number.
Now, what about ##p_2##?
Intuitive insight: the probability of getting the girl on the first choice is the same as same as getting the girl on the second choice, and the same on the third choice. This insight might not be so easy to see. Let's calculate and see:
$$p(GXX) = \frac {1} {30} \times \frac {28} {29} \times \frac {27} {28} = 0.031$$$$p(XGX) = \frac {28} {30} \times \frac {1} {29} \times \frac {27} {28} = 0.031$$$$p(XXG) = \frac {28} {30} \times \frac {27} {29} \times \frac {1} {28} = 0.031$$And we see that it's the same probability in all three sub-cases. So:
$$p_2 = p(GXX) + p(XGX) + P(XXG) = 3 \times 0.031 = 0.093$$Now, it should be clear that we have the same calculation for ##p_3##. If we do the calculution, we are simply replacing the G with a B. So:$$p_3 = 0.093$$
This now gives us a method of calculating ##p_1##, using the fact that these are mutually exclusive and represent all the possible outcomes, hence must sum to ##1##:
$$p_1 = 1 - p_2 - p_3 - p_4 = 0.007$$
Finally, it's almost always a good idea to calculate the answer two different ways and check they are the same. So, we should calculate ##p_1## directly.
Intuitive insight: all six sub-cases of getting both twins have the same probability. That is to say:
$$p(GBX) = p(GXB) = p(XGB) = p(BGX) = p(BXG) = p(XBG)$$You can check this for yourself. In any case:
$$p(GBX) = \frac{1}{30} \times \frac {1}{29} \times \frac{28}{28} = 0.00115$$$$p_1 = 6 \times p(GBX) = 0.007$$That's how I would approach the problem if I were revising probability theory. And, I can now answer any question relating to picking one or both of the twins.
