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I have a question as below: Let there be 3 groups of people, Group A (60%), B(25%), and C(15%) of the whole population. What is the expected probability that a person from Group A will randomly meet a person from Group A, Group B, and Group C, respectively?

I came up with this question in a social study I am doing recently. I'm not a mathematician, so what I thought is quite simple but I am not sure if it is correct. Please provide your opinion, and do feel free to make assumptions or conditions for solving the question if it is neccessary.

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    $\begingroup$ If you mean the next person she meets, the probabilities are $60\%$, $25\%$. $15\%$, with slight correction if the population is very small, like $12,5,3$ since someone cannot meet herself. If there are several interactions, the question should be altered, like "Among the next $4$ people she meets, what is the probability at least one is a C? $\endgroup$
    – André Nicolas
    Feb 18, 2013 at 12:40
  • $\begingroup$ Thanks very much for your responses. The answers look quite straightforward. I totally agree with the first person how the "meeting" process takes place should be specified. Hence, when it comes to the probability of 2 Groups meeting with each other (undirected), the second answer is appropriate. $\endgroup$
    – user63104
    Feb 20, 2013 at 9:29

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$$\mathbb{P}(G_1 \text{ and } G_1) = \mathbb{P}(G_1)^2$$ $$\mathbb{P}(G_2 \text{ and } G_2) = \mathbb{P}(G_2)^2$$ $$\mathbb{P}(G_3 \text{ and } G_3) = \mathbb{P}(G_3)^2$$ $$\mathbb{P}(G_1 \text{ and } G_2) = 2\mathbb{P}(G_1)\mathbb{P}(G_2)$$ $$\mathbb{P}(G_1 \text{ and } G_3) = 2\mathbb{P}(G_1)\mathbb{P}(G_3)$$ $$\mathbb{P}(G_2 \text{ and } G_3) = 2\mathbb{P}(G_2)\mathbb{P}(G_3)$$

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    $\begingroup$ You may want to edit this so it's a bit more readable. $\endgroup$
    – mrf
    Feb 18, 2013 at 15:43

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