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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)&amp;=&amp;1-P(A)-\frac{1}{2}P(A)+\frac{1}{2}[P(A)]^2\\
     &amp;=&amp; 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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