varotkorn wrote:
Q1. If this question were replacement problem, would the answer be
1/12 + 1/12 - (1/12)*(1/12) ?
Yes, you're using Venn diagram principles, and you can solve that way. But you'll have a lot more flexibility (i.e. you'll be able to solve a wider variety of problems) if you can think about this using probability principles. If we're selecting with replacement (so Joey could run all three races, in theory), then there are four possibilities for the second and third races, with the following probabilities:
Joey runs both races: (1/12)(1/12) = 1/144
Joey runs 2nd but not 3rd: (1/12)(11/12) = 11/144
Joey runs 3rd but not 2nd: (1/12)(11/12) = 11/144
Joey does not run: (11/12)(11/12) = 121/144
If you want to answer the question "what is the probability Joey runs in at least one of the second and third races?" you would add the first three probabilities above (you would not subtract one of them, as you did) to get 23/144. It's more efficient, though, to just find the probability Joey does not run at all, which is 121/144, and subtract that from 1.
varotkorn wrote:
Q2. Since the original question is NON-replacement (it's like choosing 3 positions SIMULTANEOUSLY), it is mutually exclusive (meaning that Joey can't run second and third at the same time). Hence, the (1/12)*(1/12) part disappears?
Yes, that's one way to look at it.
varotkorn wrote:
Q3. We can add together 1/12 + 1/12 ONLY when all the 12 members are DISTINCT, right?
This approach cannot be generalized to, say, 4 red marbles and 8 blue marbles where there are repeats.
Yes, that's true, because picking red second and picking red third are not mutually exclusive if you have several red marbles - both things can happen. But you could just answer this kind of question using the multiplication methods I used above (being sure to correctly account for whether the selections are made with or without replacement).
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