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I'm dealing better with probability now, but now I've faced an weird naming:

$P(\{a2, a3\}) = 2*P(a1)$

I'm guessing this {} stands for the union of both a2 and a3, am I right ?, how would I find a1 now as I don't have any value of the functions, would $P(a1) = \dfrac{P(\{a2, a3\})}{2}$ right ?

Thanks in advance;

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Have you found the "accept" button yet? –  Henry Apr 12 '12 at 0:20
honestly, not yet, where's it, and for what it stands about ? –  aajjbb Apr 12 '12 at 0:28
There is a little grayed-out check-mark under the up/down arrows at the top left corner of an answer post (underneath where you vote up or down). Click on it to accept an answer. You may want to wait a bit before accepting an answer, as a better one might be given later. –  David Mitra Apr 12 '12 at 0:33
Thanks a lot, I've already accepted your answer. –  aajjbb Apr 12 '12 at 0:46

1 Answer 1

up vote 1 down vote accepted

Yes, it is the union. Note $\{a_1,a_2,a_3\}=\{a_2,a_3\}\cup\{a_1\}$; so, assuming the only outcomes are $a_1$, $a_2$, and $a_3$ $$\tag{1} P(\{a_2,a_3\})+P(\{a_1\}) =P(\{a_1,a_2,a_3\})=1. $$ From the given information you have, you can solve the above equation for $P(a_1)$ (start by replacing $P(\{a_2,a_3\})$ in $(1)$ with $2P(a_1)$ ).

(The (incorrect) notation usually used is $P(a_1)=P(\{a_1\})$ when it won't cause confusion.)

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oh, taking it this way, so $P(a1) + P(a2) + P(a3) = 1$, so, as the tree functions has equal probability, $P(a1) = \dfrac{1}{3}$, isn't it ? –  aajjbb Apr 12 '12 at 0:24
@aajjbb It's $1/3$ but the three probabilities do not have to be equal. For instance $P(a_2)=1/2$ and $P(a_3)=1/6$ would work. All you know is that $P(\{a_2,a_3\})=2P(a_1)$. Solving equation $(1)$ above would give you $P(a_1)=1/3$. You can't say anything about $P(a_2)$ and $P(a_3)$ other than the sum of thier probabilities is $2/3$. –  David Mitra Apr 12 '12 at 0:28
oh, so clever, I didn't thought at this point, Thank you so much, + 1 for help and patience. by the way, which 'accept' button the guy over here is talking about ? –  aajjbb Apr 12 '12 at 0:34

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