Data Sufficiency

AAPL wrote:Manhattan Prep

Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?

1. The probability of selecting exactly 2 apples is greater than 1/2.
2. The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.

\[2\,\,{\text{extractions}}\,\,{\text{from}}\,\,8\,\,\left\{ \begin{gathered}
\,{\text{apples}}\,\,\left( A \right) \hfill \\
\,{\text{bananas}}\,\,\left( {B = 8 - A} \right) \hfill \\
\end{gathered} \right.\]
\[? = P\left( {{\text{both}}\,{\text{extractions}}\,\,{\text{bananas}}} \right)\]
\[\left( 1 \right)\,\,\,P\left( {{\text{both}}\,{\text{extractions}}\,\,{\text{apples}}} \right) = \frac{{C\left( {A,2} \right)}}{{C\left( {8,2} \right)}} > \frac{1}{2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,A\left( {A - 1} \right) > 28\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,A \geqslant 6\,\,\,\,\,\,\,\,\,\]
\[\left\{ {\begin{array}{*{20}{c}}
{{\text{If}}\,\,\left( {A,B} \right) = \left( {6,2} \right)} \\
{{\text{If}}\,\,\left( {A,B} \right) = \left( {7,1} \right)}
\end{array}\begin{array}{*{20}{c}}
{\,\,\, \Rightarrow \,\,\,\,\,? = \frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}} \\
{ \Rightarrow \,\,\,\,\,\,\,? = 0}
\end{array}\,\,\,\,} \right.\]
\[\left( 2 \right)\,\,\,P\left( {{\text{one}}\,\,{\text{each}}} \right) = \frac{{A \cdot \left( {8 - A} \right)}}{{C\left( {8,2} \right)}} > \frac{1}{3}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,A\left( {8 - A} \right) > 9\frac{1}{3}\,\,\,\,\,\,\,\,\,\]
\[\left\{ \begin{gathered}
\,{\text{Retake}}\,\,\,\left( {A,B} \right) = \left( {6,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}\,\, \hfill \\
\,{\text{If}}\,\,\left( {A,B} \right) = \left( {5,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \frac{{C\left( {3,2} \right)}}{{C\left( {8,2} \right)}} = \frac{3}{{28}}\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
\,A \geqslant 6 \hfill \\
\,A\left( {8 - A} \right) > 9\frac{1}{3} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 6\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{SUFF}}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,?\,\, = \,\,\frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}\,} \right]\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.