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Suppose $\Omega = \{a, b, c\}$, $P (\{a, b\}) = 0.7$, and $P (\{b, c\}) = 0.6$. Compute the probabilities of $\{a\}$, $\{b\}$, and $\{c\}$.

Is the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ useful here?

I can't seem to wrap my head around this, how can I find individual probabilities given the probability of $a$ and $b$ occurring, and $b$ and $c$ occurring?

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$P$ is a probability measure, it means $P(\Omega)=1$, and yes, the formula you mentioned is useful, apply it for $A=\{a,b\}$ and $B=\{b,c\}$ to find $P(\{b\})$, and then use $$P(\{a\})=P(\{a,b\})\ -\ P(\{b\}) \\ P(\{c\})=P(\{b,c\})\ -\ P(\{b\})$$ (by the same formula, using $\{a\}\cap\{b\}=\emptyset$ and $P(\emptyset)=0$).

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We have $P(\{ a \} ) + P(\{ b \} ) + P(\{ c \} ) = 1$. Now $$P(\{a, b \} ) = P(\{ a \} ) + P(\{ b \}) = 0.7$$ and $$P(\{ b, c \} ) = P(\{ b \} ) + P(\{ c \}) = 0.6.$$ Adding these equations yields $$P(\{ a \}) + 2P(\{ b \} ) + P(\{ c \} ) = 1.3$$. Referring to the equation in the first line we can see that $P(\{ b \} ) = 0.3$. You should be able to calculate the remaining probabilities.

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