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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, 23:20
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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, 13:04
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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


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, 13:10
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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, 21:11
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, 18:50
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



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, 07:15
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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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Brent
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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, 09:05
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)  [#permalink]

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New post 15 Apr 2019, 04:05
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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


Method 1:

Probability = Favourable outcomes / Total outcomes = Total Pairs / Total ways of picking 2 shoes out of 24

i.e. Required probability \(= \frac{12C1}{24C2} = \frac{12}{(12*23)} = \frac{1}{23}\)

Answer: Option C

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability \(= \frac{(24*1)}{(24*23)}\)

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

Answer: Option 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 16 Apr 2019, 17:55
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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 probability that the dog will drag out any matching pair is:

2/24 x 1/23 = 1/12 x 1/23

Since there are 12 different pairs, the total probability is 12 x (1/12 x 1/23) = 1/23.

Alternate Solution:

The dog can drag any shoe out, leaving 23 shoes under the bed. The probability that the next shoe he drags out is its match is 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 24 Sep 2019, 02:38
Ways to select first = 24
Ways to select second = 1
But since the order is irrelevant, there are really just 12 ways
Total ways to choose 2 from 24 (order irrelevant) = 24C2

hence 12/24C2
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 24 Sep 2019, 02:40
GMATinsight wrote:

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability \(= \frac{(24*1)}{(24*23)}\)

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

Answer: Option C


why do you not have to divide both (24*1) and (24*23) by 2? Or did you just leave that out?
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 24 Sep 2019, 19:52
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ghnlrug wrote:
GMATinsight wrote:

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability \(= \frac{(24*1)}{(24*23)}\)

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

Answer: Option C


why do you not have to divide both (24*1) and (24*23) by 2? Or did you just leave that out?



ghnlrug

In probability calculation, you must calculate both numerator and denominator either by selection (Method 1) or by Arrangement (Method 2) so in either case, you get the correct answer therefore dividing by 2 wasn't needed.
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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

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New post 24 Sep 2019, 20:01
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


Probability = 11*1/24*23 = 1/23

IMO C

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Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)   [#permalink] 24 Sep 2019, 20:01
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