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QUESTION
 A box contains 12 balls numbered from 1 to 12. The balls
   numbered 1 to 5 are red, those numbered 6 to 9 are white and
   the remaining three balls are blue. Three balls are to be drawn
   out at random without replacement from the box. Let A denote
   the event that each number drawn will be even, B the event that
   no blue ball will be drawn and C the event that one ball of
   each colour will be drawn. Calculate

    \begin{description}

     \item[(i)]$P(A)$

     \item[(ii)]$P(B)$

     \item[(iii)]$P(C)$

     \item[(iv)]$P(A\cap C)$

     \item[(v)]$P(B\cup C)$

     \item[(vi)]$P(A\cup B)$

    \end{description}

ANSWER
 \begin{description}

    \item[(i)]$A=\{\textrm{all even}\}\ \ \ \ $

     \begin{eqnarray*}
      P(A)&=&\frac{\left(\begin{array}{c}6\\3\end{array}\right)}
      {\left(\begin{array}{c}12\\3\end{array}\right)}\\&=&
      \frac{6}{12}\times\frac{5}{11}\times\frac{4}{10}
      =\frac{1}{11}
     \end{eqnarray*}

    \item[(ii)]$B=\{\textrm{no blue ball}\}\ \ \ \ $

      \begin{eqnarray*}
       P(B)&=&\frac{\left(\begin{array}{c}9\\3\end{array}\right)}
       {\left(\begin{array}{c}12\\3\end{array}\right)}\\&=&
       \frac{9}{12}\times\frac{8}{11}\times\frac{7}{10}
       =\frac{21}{55}
      \end{eqnarray*}

    \item[(iii)]$C=\{\textrm{one of each colour}\}\ \ \ \ $

      \begin{eqnarray*}
       P(C)&=&\frac{\left(\begin{array}{c}5\\1\end{array}\right)
       \left(\begin{array}{c}4\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right)}
       {\left(\begin{array}{c}12\\3\end{array}\right)}\\&=&
       \frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}\times 3!
       =\frac{3}{11}
      \end{eqnarray*}

    \item[(iv)]

      \begin{eqnarray*} P(A \cap C)&=& P(\textrm{all even}\cap \textrm{ one of
        each colour})\\ &=& \frac{\left(\begin{array}{c}2\\1\end{array}\right)
        \left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}2\\1\end{array}\right)}
        {\left(\begin{array}{c}12\\3\end{array}\right)}\\&=&
        \frac{2}{12}\times \frac{2}{11}\times \frac{2}{10}\times 3!
        =\frac{2}{55}
      \end{eqnarray*}

    \item[(v)]$P(B \cup C)=P(B)+P(C)-P(B \cap C)$ by addition
     theorem.$P(B \cap C)=P(\textrm{no blue} \cap \textrm{one of
     each colour}=0 \\
     \textrm{therefore} P(B \cup C)=P(B)+P(C)=\frac{36}{55}$

    \item[(vi)]$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ by addition
     theorem.$P(A \cap B)=P(\textrm{all even} \cap \textrm{no blue}=
     \frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}=\frac{1}{55} \\
     \textrm{therefore} P(A \cup B)=\frac{1}{11}+\frac{21}{55}-\frac{1}{55}=\frac{36}{55}$

   \end{description}

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