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Numerator: Out of 20 shoes we have to select one shoe first..w/o any restrictions.So that's 20C1 (choosing one out of 20). Now,the second show has to be the matching shoe.There is only ONE such possible shoe,so the probability of choosing that shoe is 1.

Re: probability that they are matching shoes? [#permalink]
28 Oct 2009, 00:39

1

This post received KUDOS

Expert's post

tejal777 wrote:

A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes? 1/190

Please see my method: ...................................................

Numerator: Out of 20 shoes we have to select one shoe first..w/o any restrictions.So that's 20C1 (choosing one out of 20). Now,the second show has to be the matching shoe.There is only ONE such possible shoe,so the probability of choosing that shoe is 1.

Denominator: Chosing 2 shoes out of 20.So, 20C2

20C1/20C2 = 2/19

please explain:(

The problem with your solution is that we don't choose 1 shoe from 20, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/19(as after taking one at random there are 19 shoes left and only one is the pair of the first one)=1/19

Answer: C.

We can solve it like you were doing:

P=Favorable outcomes/Total # of outcomes

Favorable outcomes = 10C1 as there are 10 pairs and we need ONE from these 10 pairs. Total # of outcomes= 20C2 as there are 20 shoes and we are taking 2 from them.

Re: probability that they are matching shoes? [#permalink]
28 Oct 2009, 00:59

Hey Bunuel thanks for the quick reply!However I am still not clear.Here is how our solution differs, You are saying P=10C1/20C2 While in my solution it is 20C1/20C2

Can you please clarify this difference?Shoul'nt we be considering the answer wrt either pairs or total no. of shoes?in your answer in the numerator you are selecting one out of 10 PAIRS while I am selecting 1 out of 20 shoes..lol..is'nt it the same thing?(apparently not!) _________________

Re: probability that they are matching shoes? [#permalink]
28 Oct 2009, 01:26

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

tejal777 wrote:

Hey Bunuel thanks for the quick reply!However I am still not clear.Here is how our solution differs, You are saying P=10C1/20C2 While in my solution it is 20C1/20C2

Can you please clarify this difference?Shoul'nt we be considering the answer wrt either pairs or total no. of shoes?in your answer in the numerator you are selecting one out of 10 PAIRS while I am selecting 1 out of 20 shoes..lol..is'nt it the same thing?(apparently not!)

You are right it's not the same thing:

First of all: In numerator you have the number of selection of ONE shoe out of 20, but in denominator TWO shoes out of 20. In my solution both numerator and denominator counting the # of selections of TWO shoes, though from different quantities.

Second, consider this: what is your winning (favorable) probability? You want to choose ONE pair. Out of how many pairs? Out of 10. Now, what are you actually doing? You are taking TWO shoes out of 20, "hoping" that this two will be the pair. So anyway wining scenarios are 10C2 and total number of scenarios are 20C2.

You can do this also in the following way (maybe this one would be easier to understand): Numerator: 20C1 - # of selection of 1 shoe out of 20, multiplied by 1C1, as the second one can be only ONE, as there is only one pair of chosen one. Which means that # of selection would be 1C1. Put this in numerator.

Denominator would be: 20C1 # of selection of 1 from 20, multiplied be 19C1 # of selection from the 19 left.

So, P=20C1*1C1/20C1*19C1=1/19, here we have in the numerator the same thing you wrote, BUT if we are doing this way then in denominator you should also count the # of selection of the first one out of 20 and the second one out of 19.

The first way of solving actually is the same. Take one shoe from 20, any shoe from 20, I mean just randomly take one. Then you are looking at your 19 left shoes and want to choose the pair of the one you've already taken, as in 19 you have only one which is the pair of the first one you have the probability 1/19 (1 chance out of 19) to choose the right one. _________________

Re: probability that they are matching shoes? [#permalink]
28 Oct 2009, 08:47

I did it this way

1/10 * 1/19 * 10 = 1/19

Reasoning: The first shoe you pick is one of the pair out of 10 pairs so it is 1/10 The second shoe has to match the first shoe so it is now 1 out of 19 remain shoes (not pairs) There are 10 pairs so the probably of the first 2 events times 10

Re: probability that they are matching shoes? [#permalink]
06 Apr 2011, 01:41

jzd, your approach is different. cos if i am already selecting 1 out of 10 pairs, that 1 corresponds to 1 pair and not once piece of shoe, if so then why would i again pick another shoe out of balance 19?pls explain.bunuel explanation is more effective it seems... correct me if i am wrong _________________

"When the going gets tough, the tough gets going!"

Re: probability that they are matching shoes? [#permalink]
17 Oct 2012, 06:54

just wondering what would be the answer if all 10 pairs were of same size.

In fact, I do not know why, but I inadvertently assumed that they are of same size and arrived at the answer 10/19, which I obviously did not find in the choices. Help please!

Re: probability that they are matching shoes? [#permalink]
17 Oct 2012, 07:39

1

This post received KUDOS

Expert's post

anilisanil wrote:

just wondering what would be the answer if all 10 pairs were of same size.

In fact, I do not know why, but I inadvertently assumed that they are of same size and arrived at the answer 10/19, which I obviously did not find in the choices. Help please!

Size of the shoes is not the main point here. We are not told that these 10 pairs are identical (size, shape, color, ...) thus we cannot assume that.

Next, if we were told that there are 10 identical pairs of shoes, then the answer would be 10/19: after we choose one shoe, then there are 10 matching shoes left there out of 19, so P=10/19.

Re: probability that they are matching shoes? [#permalink]
17 Oct 2012, 07:44

1

This post received KUDOS

anilisanil wrote:

just wondering what would be the answer if all 10 pairs were of same size.

In fact, I do not know why, but I inadvertently assumed that they are of same size and arrived at the answer 10/19, which I obviously did not find in the choices. Help please!

You computed the probability of choosing a pair of shoes, one left and one right shoe, from 10 identical pairs. The probability to choose the first shoe is 20/20 = 1 (it can be any one of them). The probability to choose a matching shoe for the first one (meaning choose a left shoe after a right shoe, or the other way around) is 10/19.

In our case, when we have 10 different pairs of shoes, the probability to choose the first shoe is 20/20 = 1, then the probability to choose the matching shoe for the previous one is 1/19. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A box contains 10 pairs of shoes (20 shoes in total). If two [#permalink]
17 Jul 2013, 01:33

1

This post received KUDOS

probability to choose any shoe out of 20 is 20/20 probability to choose its pair is 1 out of 19 i.e 1/19.

thus (20/20)*(1/19) = 1/19 _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: A box contains 10 pairs of shoes (20 shoes in total). If two [#permalink]
26 Jan 2014, 11:06

Consider {L1, R1}, {L2, R2}.....{L10, R10}.

We can pick any of the left ones. probability = 1/10 (let's say we have picked L1). The probability of picking R1 = 1/19. Probability of picking L1 and R1 is 1/10 * 1/19 = 1/190.

However, we can pick {L2, R2}, {L3, R3}.....{L10, R10}. So, required probability = 10 * 1/190 = 1/19

Re: probability that they are matching shoes? [#permalink]
20 Apr 2014, 15:10

Bunuel wrote:

tejal777 wrote:

Hey Bunuel thanks for the quick reply!However I am still not clear.Here is how our solution differs, You are saying P=10C1/20C2 While in my solution it is 20C1/20C2

Can you please clarify this difference?Shoul'nt we be considering the answer wrt either pairs or total no. of shoes?in your answer in the numerator you are selecting one out of 10 PAIRS while I am selecting 1 out of 20 shoes..lol..is'nt it the same thing?(apparently not!)

You are right it's not the same thing:

First of all: In numerator you have the number of selection of ONE shoe out of 20, but in denominator TWO shoes out of 20. In my solution both numerator and denominator counting the # of selections of TWO shoes, though from different quantities.

Second, consider this: what is your winning (favorable) probability? You want to choose ONE pair. Out of how many pairs? Out of 10. Now, what are you actually doing? You are taking TWO shoes out of 20, "hoping" that this two will be the pair. So anyway wining scenarios are 10C2 and total number of scenarios are 20C2.

You can do this also in the following way (maybe this one would be easier to understand): Numerator: 20C1 - # of selection of 1 shoe out of 20, multiplied by 1C1, as the second one can be only ONE, as there is only one pair of chosen one. Which means that # of selection would be 1C1. Put this in numerator.

Denominator would be: 20C1 # of selection of 1 from 20, multiplied be 19C1 # of selection from the 19 left.

So, P=20C1*1C1/20C1*19C1=1/19, here we have in the numerator the same thing you wrote, BUT if we are doing this way then in denominator you should also count the # of selection of the first one out of 20 and the second one out of 19.

The first way of solving actually is the same. Take one shoe from 20, any shoe from 20, I mean just randomly take one. Then you are looking at your 19 left shoes and want to choose the pair of the one you've already taken, as in 19 you have only one which is the pair of the first one you have the probability 1/19 (1 chance out of 19) to choose the right one.

Hi Bunuel,

Can you talk about why, in your second example you have to break down the denominators into 20c1 AND 19c1 instead of just 20c2?

A box contains 10 pairs of shoes (20 shoes in total). If two [#permalink]
12 Sep 2014, 07:30

Bunuel wrote:

tejal777 wrote:

A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes? 1/190

Please see my method: ...................................................

Numerator: Out of 20 shoes we have to select one shoe first..w/o any restrictions.So that's 20C1 (choosing one out of 20). Now,the second show has to be the matching shoe.There is only ONE such possible shoe,so the probability of choosing that shoe is 1.

Denominator: Chosing 2 shoes out of 20.So, 20C2

20C1/20C2 = 2/19

please explain:(

The problem with your solution is that we don't choose 1 shoe from 20, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/19(as after taking one at random there are 19 shoes left and only one is the pair of the first one)=1/19

Answer: C.

We can solve it like you were doing:

P=Favorable outcomes/Total # of outcomes

Favorable outcomes = 10C1 as there are 10 pairs and we need ONE from these 10 pairs. Total # of outcomes= 20C2 as there are 20 shoes and we are taking 2 from them.

P=10C1/20C2 =10/(19*10)=1/19

Answer: C.

My apologies if this is already somewhere on the forum, but I've searched and can't figure out any of the combination explanations because everyone always skips the last step. Can you please explain the math that went behind changing 10C1/20C2 into 1/19? I'm guessing this means you had to use the formula from the GMAT Book you all made?

So, I'm guessing this means... \frac{10!}{1!9!} divided by \frac{20!}{2!18!}

If so, how did you quickly calculate the factorials like that? Is there a shortcut I'm missing?

Last edited by Chin926926 on 12 Sep 2014, 07:37, edited 1 time in total.

Re: A box contains 10 pairs of shoes (20 shoes in total). If two [#permalink]
12 Sep 2014, 07:35

Expert's post

1

This post was BOOKMARKED

Chin926926 wrote:

Bunuel wrote:

tejal777 wrote:

A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes? 1/190

Please see my method: ...................................................

Numerator: Out of 20 shoes we have to select one shoe first..w/o any restrictions.So that's 20C1 (choosing one out of 20). Now,the second show has to be the matching shoe.There is only ONE such possible shoe,so the probability of choosing that shoe is 1.

Denominator: Chosing 2 shoes out of 20.So, 20C2

20C1/20C2 = 2/19

please explain:(

The problem with your solution is that we don't choose 1 shoe from 20, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/19(as after taking one at random there are 19 shoes left and only one is the pair of the first one)=1/19

Answer: C.

We can solve it like you were doing:

P=Favorable outcomes/Total # of outcomes

Favorable outcomes = 10C1 as there are 10 pairs and we need ONE from these 10 pairs. Total # of outcomes= 20C2 as there are 20 shoes and we are taking 2 from them.

P=10C1/20C2 =10/(19*10)=1/19

Answer: C.

Can you please explain the math that went behind changing 10C1/20C2 into 1/19? I'm guessing this means you had to use the formula from the GMAT Book you all made?

So, I'm guessing this means... \frac{10!}{1!9!} divided by \frac{20!}{2!18!}

If so, how did you quickly calculate the factorials like that? Is there a shortcut I'm missing?

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...