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Emily keeps 12 different pairs of shoes (24 individual shoes in total)

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Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 16 Sep 2018, 22:20
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Question Stats:

70% (01:03) correct 30% (01:15) wrong based on 132 sessions

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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

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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 17 Sep 2018, 12:04
Bunuel wrote:
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


No of ways selecting the 1st shoe = 24
No of ways selecting the 1st & 2nd shoe= 24 *23
No of ways selecting the 2nd shoe which is the exact pair of the 1st = 1
No of ways selecting selecting the 1st shoe & the 2nd shoe which is the exact pair of the 1st= 24*1
Hence P = 24*1 /24*23 = 1/23 ....... Ans C
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 17 Sep 2018, 12:10
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Probability of the first one don't not matter, so we should consider what is the probability of the second one be the pair of the first one.

1/23

Answer C
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 17 Sep 2018, 20:11
Bunuel wrote:
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



Hi,

My approach-

Probability of selecting a particular shoe from the 12 different pair=2/24
Now probability of selecting the matching shoe to the shoe already dragged= 1/23(remember one shoe is already out)

Therefore combined probability of one matching pair of shoe= 1/24* 1/23

Now, there are 12 different pair of shoe so your probability will shoot up 12 times-

2/24 *1/23 *12=1/23.

Bunuel is it correct?
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 18 Sep 2018, 17:50
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Bunuel wrote:
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



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 of the two shoes forming a matching pair is 1 x 1/23 = 1/23.

Answer: C
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 09 Jan 2019, 06:15
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Bunuel wrote:
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

Answer: C

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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 09 Jan 2019, 08:05
1
Bunuel wrote:
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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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total) &nbs [#permalink] 09 Jan 2019, 08:05
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Emily keeps 12 different pairs of shoes (24 individual shoes in total)

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