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## Emily keeps 12 different pairs of shoes (24 individual shoes

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### Emily keeps 12 different pairs of shoes (24 individual shoes

by AAPL » Wed Jan 09, 2019 5:03 am

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## Global Stats

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Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

OA C

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by Brent@GMATPrepNow » Wed Jan 09, 2019 6:05 am
AAPL wrote:Veritas Prep

Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11
P(dog selects matching pair) = P(dog chooses ANY sock 1st AND 2nd sock matches the 1st sock)
= P(dog chooses ANY sock 1st) x P(2nd sock matches the 1st sock)
= 24/24 x 1/23
= 1/23

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by fskilnik@GMATH » Wed Jan 09, 2019 7:54 am
AAPL wrote:Veritas Prep

Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11
$$\left. \matrix{ {\rm{Total}}\,\,:\,\,\,C\left( {24,2} \right) = {{24 \cdot 23} \over 2} = 12 \cdot 23\,\,{\rm{equiprobable}}\,\,{\rm{pairs}}\,\,\,\, \hfill \cr {\rm{Favorable:}}\,\,\,12\,\,{\rm{real}}\,\,{\rm{pairs}}\,\,\left( {{\rm{matches}}} \right) \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {{12} \over {12 \cdot 23}} = {1 \over {23}}$$

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

Regards,
Fabio.
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by Scott@TargetTestPrep » Mon Jan 21, 2019 6:00 pm
AAPL wrote:Veritas Prep

Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

OA C
The first shoe can be any shoe, so its probability is 24/24 = 1. However, since the second shoe must match the first shoe, its probability of being chosen is 1/23. Therefore, the probability that the two shoes forming a matching pair is 1 x 1/23 = 1/23.