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QUESTION

 Two events A and B are such that the probability of B given A is
   4 times the probability of A and the probability of A given B
   is 9 times the probability of B. If the probability that at
   least one of A and B occurs is $7/48$, Find the probability of
   A.

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ANSWER

$P(B|A)=4P(A)\rightarrow \frac{P(A \cap B}{P(A)}=4P(A)\\
   P(A|B)=9P(B)\rightarrow \frac{P(A \cap B}{P(B)}=9P(B)\\
   \textrm{therefore }P(A\cap B)=4[P(A)]^2=9[P(B)]^2\\
   \textrm{therefore } 2P(A)=3P(B)$\\

    \begin{eqnarray*}
     \frac{7}{48}=P(A\cup B)&=&P(A)+P(B)-P(A \cap B)\\
     &=&P(A)+\frac{2}{3}P(A)-4[P(A)]^2
    \end{eqnarray*}

   $$192[P(A)]^2-80P(A)+7=0\\
   (8P(A)-1)(24P(A)-7)=0$$
   $P(A)=\frac{1}{8} \textrm{ or }P(A)=\frac{7}{24}. \textrm{Since
   }P(A)=\frac{7}{24} >\frac{7}{48}=P(A \cup B)$ \  this is not a
   possible solution. Therefore $P(A)=\frac{1}{8}$


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