gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?
(1) The probability that one ball selected at random is either red or white is \(\frac{2}{3}\)
(2) There are nine balls in the barrel.
\(?\,\,\,\, = \,\,\,P\left( {2\,\,br\,\,{\text{out}}\,\,{\text{of}}\,\,2\,\,{\text{sequential}}\,\,{\text{extractions}}\,\,{\text{no}}\,\,{\text{replacements}}} \right)\,\,\)
\(\left( 1 \right)\,\,P\left( {re\,\,{\text{or}}\,\,wh\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{2}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,P\left( {br\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{1}{3}\)
\(\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {2,2,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,r + w + b\,\, = \,\,9\)
\(\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {7,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {6,1,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,br = \frac{1}{3}\left( 9 \right) = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{\text{unique}}\,\,\,\,\left( {\frac{3}{9} \cdot \frac{2}{8}} \right)\)
The above follows the notations and rationale taught in the GMATH method.
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