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Let $A$ and $B$ be two independent events such that $P(B)>P(A)$. If the probability that both A and B happen is $\frac{1}{12}$ and the probability that neither Anor B happens is $\frac{1}{2}$ then
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The correct answer is:
$\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{3}$
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