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QUESTION
A and B are two independent events. A is twice as likely to
occur as B and three times as likely to occur as the event that
neither A or B does. Find the probability of A.
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ANSWER
A and B are independent therefore $P(A \cap B)=P(A)P(B)$\\
A is twice as likely as B therefore $P(A)=2P(B)$\\
A is three times as likely as neither A or B therefore
$P(A)=3P(\overline{A} \cap \overline{B})$\\
$P(\overline{A} \cap \overline{B})1-P(A\cup B)=1-P(A)-P(B)-P(A\cap B)$\\
\begin{eqnarray*}\frac{1}{3}P(A)&=&1-P(A)-\frac{1}{2}P(A)+\frac{1}{2}[P(A)]^2\\
&=& 1-\frac{3}{2}P(A)+\frac{1}{2}[P(A)]^2
\end{eqnarray*}
$$3[P(A)]^2-11P(A)+6=0 $$
$$(3P(A)-2)(P(A)-3)=0 $$\\
$P(A)=\frac{2}{3}\textrm{ or }P(A)=3. \textrm{ Since }0\leq
P(A)\leq 1\ \ \ \ P(A)=\frac{2}{3}$
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