Two pieces of fruit are selected out of a group of 8 pieces

Choice (2) makes it clear that there is banana in the group of fruits, doesn't it? And yeah, it's always bad to assume ANYTHING on gmat, especially for Data Sufficiency and CR questions! So, when considering choice (1) by itself, no. of bananas=0 should also be one of the options.

Hello Bunuel,

I have three questions:

1. The question states, "8 pieces of fruit consisting only of apples and bananas," does this not mean the number of bananas cannot be 0?

2. Since the phrase, "8 pieces of fruit consisting only of apples and bananas," writes bananas, a plural, should this mean that the number of bananas would be greater than 1?

3. Also, the question asks the probability of getting exactly 2 bananas, so doesn't that imply that there are at least 2 bananas?

Just concerned. When the question statement says that there are bananas AND apples, do we need to consider situations in which there are only apples or only bananas??? I'm asking this not for just this question but for the GMAT on the whole.

I have solved this question with similar logic, but answered E because i understoon the 2nd statement as no matter what is the order the probability will be greater than 1/3, but in your solution i see that "in either order" means in both ways. Could you please clarify that?

The probability of selecting 1 apple and 1 banana in either order equals to the probability of selecting an apple and then a banana (x/8*(8-x)/7) PLUS the probability of selecting a banana and then an apple ((x-8)/8*x/7) --> x/8*(8-x)/7+(8-x)/8*x/7=2*x/8*(8-x)/7.

Hope it's clear.

\(2\,\,{\text{extractions}}\,\,{\text{from}}\,\,8\,\,\left\{ \begin{gathered}\\
\,{\text{apples}}\,\,\left( A \right) \hfill \\\\
\,{\text{bananas}}\,\,\left( {B = 8 - A} \right) \hfill \\ \\
\end{gathered} \right.\)

\(? = P\left( {{\text{both}}\,{\text{extractions}}\,\,{\text{bananas}}} \right)\)


\(\left( 1 \right)\,\,\,P\left( {{\text{both}}\,{\text{extractions}}\,\,{\text{apples}}} \right) = \frac{{C\left( {A,2} \right)}}{{C\left( {8,2} \right)}} > \frac{1}{2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,A\left( {A - 1} \right) > 28\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,A \geqslant 6\,\,\,\,\,\,\,\,\,\)

\(\left\{ {\begin{array}{*{20}{c}}\\
{{\text{If}}\,\,\left( {A,B} \right) = \left( {6,2} \right)} \\ \\
{{\text{If}}\,\,\left( {A,B} \right) = \left( {7,1} \right)} \\
\end{array}\begin{array}{*{20}{c}}\\
{\,\,\, \Rightarrow \,\,\,\,\,? = \frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}} \\ \\
{ \Rightarrow \,\,\,\,\,? = 0} \\
\end{array}\,\,\,\,} \right.\)


\(\left( 2 \right)\,\,\,P\left( {{\text{one}}\,\,{\text{each}}} \right) = \frac{{A \cdot \left( {8 - A} \right)}}{{C\left( {8,2} \right)}} > \frac{1}{3}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,A\left( {8 - A} \right) > 9\frac{1}{3}\,\,\,\,\,\,\,\,\,\)

\(\left\{ \begin{gathered}\\
\,{\text{Retake}}\,\,\,\left( {A,B} \right) = \left( {6,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}\,\, \hfill \\\\
\,{\text{If}}\,\,\left( {A,B} \right) = \left( {5,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \frac{{C\left( {3,2} \right)}}{{C\left( {8,2} \right)}} = \frac{3}{{28}}\,\,\, \hfill \\ \\
\end{gathered} \right.\,\,\)


\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}\\
\,A \geqslant 6 \hfill \\\\
\,A\left( {8 - A} \right) > 9\frac{1}{3} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 6\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{SUFF}}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,?\,\, = \,\,\frac{1}{{C\left( {8,2} \right)}} = \frac{1}{{28}}\,} \right]\)


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

Regards,
Fabio.

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