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Six distinct pair of shoes are in a closet. If three shoes are

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VyshakhR1995
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Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   Updated on: 23 Apr 2017, 02:57
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Six distinct pair of shoes are in a closet. If three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

A. 1/5
B. 2/11
C. 13/33
D. 8/11
E. 4/5
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D

Originally posted by VyshakhR1995 on 23 Apr 2017, 00:29.
Last edited by Bunuel on 23 Apr 2017, 02:57, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Six distinct pair of shoes are in a closet [#permalink] New post   23 Apr 2017, 02:58
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VyshakhR1995 wrote:
Six distinct pair of shoes are in a closet.if three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

1/5
2/11
13/33
8/11
4/5


Hi,
Please post Q in correct format with choices marked as A,B....

There are 6distinct pairs, so 12 shoes in all...
Way to choose 3 out of 12 =12*11*10..
Way to choose all 3 from different pairs:-
1) first can be any of 12.....So 12 ways
2) second can be any of remaining except the pair of 1st choosen, do 11-1=10 ways..
3) third can be any of remaining 10, except of pairs of first two..., So 10-2*1=8..

Probability=\(\frac{12*10*8}{12*11*10}=\frac{8}{11}\)
D
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   19 Jul 2017, 16:00
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There are 12 shoes, in 6 pairs. It doesn't matter what shoe we pick first. There will then be 11 shoes left, and 10 of those don't match our first shoe. So the probability our second selection doesn't make a pair is 10/11. Then we pick a third shoe from the 10 that remain, and there are 8 shoes left that won't make a pair (we need to pick a shoe from one of the four unused pairs of shoes that remain). We need both those things to happen, so we multiply the two probabilities: the answer is (10/11)*(8/10) = 8/11
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   12 Sep 2017, 04:35
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VyshakhR1995 wrote:
Six distinct pair of shoes are in a closet. If three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

A. 1/5
B. 2/11
C. 13/33
D. 8/11
E. 4/5


I use method 1 - opposite probability

Select 3 shoes with 2 are a pair:

Ways to select a pair: 6C1
Ways to select any shoe from 10 shoes: 10C1
Total ways to choose 3 shoes: 12C3

Probability: (6C1X10C1)/12C3 = 3/11

Probability of no pair of shoes in any three shoes: 1-(3/11)=8/11

Answer is D
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   12 Sep 2017, 03:08
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sasidharrs wrote:
Hi,
1st selection probability is 1 (choose anyone)
2nd selection probability is 1/10( since we should not select same pair of shoe)
3rd selection probability is 1/8(since we should not select the first 2 pair color shoes)
total probability : 1*1/10*1/8 = 1/80


But, my answer is wrong, I do not know why.. basically, I do am not able to decide when to use combination approach?


Your approach is almost there - the bit you are missing is thinking about the question as odds it is NOT a pair.

1st selection probability is correctly identified as 1
2nd selection probability is (12-1=11 shoes left, of which 1 is the same pair): 10/11
3rd selcetion probability is (11-1=10 shoes left, of which 2 are the same pair): 8/10

8/10*10/11 = 80/110 = 8/11

Thinking & maths was right, but problem approach was the wrong way around! Hope that helped.
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   21 Jun 2017, 20:44
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probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.
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Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   22 Jun 2017, 15:06
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VyshakhR1995 wrote:
pradeepmaria wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.


Didn't get your question.Can you explain??



12C3

\(\frac{12*11*10}{3*2*1}\)=220

This is what is meant. Why don't you divide by 6?
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   22 Jun 2017, 12:28
pradeepmaria wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.


Didn't get your question.Can you explain??
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   24 Jun 2017, 22:01
Smokeybear00 wrote:
VyshakhR1995 wrote:
pradeepmaria wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.


Didn't get your question.Can you explain??



12C3

\(\frac{12*11*10}{3*2*1}\)=220

This is what is meant. Why don't you divide by 6?


Now i think i understand the original question,but now i am confused by your's.
12*11*10 gives the arrangement.
If A B C are three items when considering arrangement we account for ACB BCA etc
For Combination we Don't need to worry about that.
If you are selecting a committee,it doesn't matter whether you Choose A or B first.But when you are arranging them you need to consider the order.
The formula you mentioned calculates the arrangement.
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   19 Jul 2017, 15:56
This question needs a detailed explanation involving multiple approaches. Can somebody help here?
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   22 Jul 2017, 14:50
Hi,
1st selection probability is 1 (choose anyone)
2nd selection probability is 1/10( since we should not select same pair of shoe)
3rd selection probability is 1/8(since we should not select the first 2 pair color shoes)
total probability : 1*1/10*1/8 = 1/80


But, my answer is wrong, I do not know why.. basically, I do am not able to decide when to use combination approach?
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   02 Jan 2019, 22:42
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pradeepmaria wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.


Hi Pradeep,

Total number of ways of selecting 3 items from 12 is indeed \(12C3\)= 220.

This is the denominator while calculating Probability

To find the number of ways in which no two shoes are from the same pair

First Shoe : 12 ways (any of the 12 shoes)
Second Shoe : 10 ways (any of the 10 shoes remaining after excluding the shoe remaining from the first pair)
Third Shoe : 8 ways

Total = 12 * 10 * 8 ways = 960

But this gives us the number of arrangements (where order matters)

To get the number of selections , we divide 960 by 3! to get 160

Required Probability = \(\frac{160}{220}\) = 8/11
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   13 Feb 2021, 22:22
Given: Six distinct pair of shoes are in a closet.
Asked:If three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

Total number of ways to select 3 shoes out of 12 = 12C3
Number of ways to select 3 shoes having no pair = 12×10×8/3!=160
Probability = 160/220 = 8/11

IMO D

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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   19 Feb 2021, 18:28
(12/12) (10/11) (8/10) = 8/11
1st shoe 2nd shoe 3rd

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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   19 Feb 2021, 18:28
(12/12) (10/11) (8/10) = 8/11
1st shoe 2nd shoe 3rd

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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink] New post   25 Jan 2024, 00:10
The three shoes chosen will come from 3 different pairs. Thus, we first need to select 3 pairs of shoes. This can be done in 6C3 ways = 20.

After we select the 3 pairs, we have to select only one shoe out of each pair - so that no two shoes make a pair. This can be done in 2C1 * 2C1 * 2C1 ways, which is equal to 2*2*2 = 8.

Thus, all favorable possibilities = 20*8 = 160.

Total possibilities = 12C3 (any 3 shoes can be chosen from 12 shoes, whether they make a pair or not). This would be equal to 220.

Thus probability = 160/220 = 8/11.
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Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink]
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