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Is a probability function in $\Omega{(a1, a2, a3)}$, find $P(a1)$ whether $P({a2, a3}) = 2P(a1)$.

I know that $1 = a1 + a2 + a3$.

from where I have to start ?

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Do you mean to say that the sample space is a three element set? That is $\Omega=\{a_1,a_2,a_3\}$? –  user21436 Apr 4 '12 at 12:48
    
This is exactly, the sample space is Ω={a1,a2,a3}. –  mastergoo Apr 4 '12 at 12:50
    
You have to start with a comprehensible question. What does $P(a_2,a_3)$ mean? –  Gerry Myerson Apr 4 '12 at 12:52
    
$a2 \Cup a3$, this is ? –  mastergoo Apr 4 '12 at 12:53
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1 Answer

up vote 3 down vote accepted

Start by using additivity: $P(\{a_2, a_3\} \cup \{a_1\}) = P(\{a_2,a_3\})+P(\{a_1\})$. Since $\{a_2, a_3\} \cup \{a_1\} = \Omega$, and $P(\Omega) = 1$, we have found that $$ P(\{a_2,a_3\})+P(\{a_1\}) = 1.$$ Now substitute $P(\{a_2,a_3\}) = 2P(\{a_1\})$ into this equation, and you'll find $P(\{a_1\}) = \frac{1}{3}$.

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Oh yes, I've find the following result: $1 = a1 + a2 + a3$, so $1 = a1 + (2a1)$, so $a1 = \frac{1}{3}$ –  mastergoo Apr 4 '12 at 12:59
    
THANK YOU SO MUCH. Very good explanation. –  mastergoo Apr 4 '12 at 13:02
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You're welcome, but you really need to be more precise with your notation: $a_1$ is the event (or rather $\{a_1\}$ is), which is very different from its probability $P(\{a_1\})$. –  lazyhaze Apr 4 '12 at 13:10
    
By the way, don't put "Solved" in the title, but instead accept my answer (if you find it acceptable, of course). –  lazyhaze Apr 4 '12 at 13:11
    
Ok ok, I undertood.. –  mastergoo Apr 4 '12 at 13:23
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