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

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Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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09 Nov 2012, 00:34
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40% (02:03) correct 60% (02:25) wrong based on 297 sessions

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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.

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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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09 Nov 2012, 02:43
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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?

Say there are $$x$$ bananas and $$y$$ ($$y=8-x$$) apples. The question is $$P(bb)=\frac{x}{8}*\frac{x-1}{7}=?$$. Basically we need to find how many bananas are there.

(1) The probability of selecting exactly 2 apples is greater than 1/2 --> $$\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}$$ --> $$y(y-1)>28$$ --> $$y$$ can be 6, 7, or 8, thus $$x$$ can be 2, 1, or 0. not sufficient.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. $$2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}$$ --> $$x(8-x)>\frac{28}{3}=9\frac{1}{3}$$, thus $$x$$ can be 2, 3, 4, 5, or 6. Not sufficient.

(1)+(2) From above x can only be 2. Sufficient.

Hope it's clear.
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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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09 Nov 2012, 02:27
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Pansi wrote:
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 ½.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3

Total No. of ways of selecting = 8C2 = 28

1)No. of apples = 7, No. of bananas = 1, Probability : $$\frac{7C2}{8C2} = \frac{21}{28}$$
No. of apples = 6, No. of bananas = 2, Probability : $$\frac{6C2}{8C2} = \frac{15}{28}$$
No. of apples = 5, No. of bananas = 3, Probability : $$\frac{5C2}{8C2} = \frac{10}{28}$$

So, No. of apples can be 6 or 7. Insufficient

2)No. of apples = 7, No. of bananas = 1, Probability : $$\frac{7C1*1C1}{8C2} = \frac{7}{28}$$
No. of apples = 6, No. of bananas = 2, Probability : $$\frac{6C1*2C1}{8C2} = \frac{12}{28}$$
No. of apples = 5, No. of bananas = 3, Probability : $$\frac{5C1*3C1}{8C2} = \frac{15}{28}$$

So, No. of apples can be 5 or 6. ( More values other than 7 are also possible, but two values are enough to make the statement insufficient.)Insufficient

1 & 2 together. No. Of apples = 6, No. of bananas = 2. Enough info to find what is required. Sufficient.

Although, I'm not very strong at combinatronics and hence I'm not 100% sure of my method. Also since the question states that there ARE bananas, I'm assuming that no. of apples cannot be 8.

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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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09 Nov 2012, 02:54
Bunuel wrote:
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?

Say there are $$x$$ bananas and $$y$$ ($$y=8-x$$) apples. The question is $$P(bb)=\frac{x}{8}*\frac{x-1}{7}=?$$. Basically we need to find how many bananas are there.

(1) The probability of selecting exactly 2 apples is greater than 1/2 --> $$\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}$$ --> $$y(y-1)>28$$ --> $$y$$ can be 6, 7, or 8, thus $$x$$ can be 2, 1, or 0. not sufficient.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. $$2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}$$ --> $$x(8-x)>\frac{28}{3}=9\frac{1}{3}$$, thus $$x$$ can be 2, 3, 4, 5, or 6. Not sufficient.

(1)+(2) From above x can only be 2. Sufficient.

Hope it's clear.

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.
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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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09 Nov 2012, 03:04
MacFauz wrote:
Bunuel wrote:
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?

Say there are $$x$$ bananas and $$y$$ ($$y=8-x$$) apples. The question is $$P(bb)=\frac{x}{8}*\frac{x-1}{7}=?$$. Basically we need to find how many bananas are there.

(1) The probability of selecting exactly 2 apples is greater than 1/2 --> $$\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}$$ --> $$y(y-1)>28$$ --> $$y$$ can be 6, 7, or 8, thus $$x$$ can be 2, 1, or 0. not sufficient.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. $$2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}$$ --> $$x(8-x)>\frac{28}{3}=9\frac{1}{3}$$, thus $$x$$ can be 2, 3, 4, 5, or 6. Not sufficient.

(1)+(2) From above x can only be 2. Sufficient.

Hope it's clear.

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.

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.
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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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11 Nov 2012, 01:53
Bunuel wrote:
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?

Say there are $$x$$ bananas and $$y$$ ($$y=8-x$$) apples. The question is $$P(bb)=\frac{x}{8}*\frac{x-1}{7}=?$$. Basically we need to find how many bananas are there.

(1) The probability of selecting exactly 2 apples is greater than 1/2 --> $$\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}$$ --> $$y(y-1)>28$$ --> $$y$$ can be 6, 7, or 8, thus $$x$$ can be 2, 1, or 0. not sufficient.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. $$2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}$$ --> $$x(8-x)>\frac{28}{3}=9\frac{1}{3}$$, thus $$x$$ can be 2, 3, 4, 5, or 6. Not sufficient.

(1)+(2) From above x can only be 2. Sufficient.

Hope it's clear.

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?
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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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12 Nov 2012, 10:00
ziko wrote:
Bunuel wrote:
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?

Say there are $$x$$ bananas and $$y$$ ($$y=8-x$$) apples. The question is $$P(bb)=\frac{x}{8}*\frac{x-1}{7}=?$$. Basically we need to find how many bananas are there.

(1) The probability of selecting exactly 2 apples is greater than 1/2 --> $$\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}$$ --> $$y(y-1)>28$$ --> $$y$$ can be 6, 7, or 8, thus $$x$$ can be 2, 1, or 0. not sufficient.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. $$2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}$$ --> $$x(8-x)>\frac{28}{3}=9\frac{1}{3}$$, thus $$x$$ can be 2, 3, 4, 5, or 6. Not sufficient.

(1)+(2) From above x can only be 2. Sufficient.

Hope it's clear.

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.
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Re: Two pieces of fruit are selected out of a group of 8 pieces  [#permalink]

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28 Oct 2018, 14:02
Pansi wrote:
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.
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