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Six distinct pair of shoes are in a closet. If three shoes are [#permalink]

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23 Apr 2017, 00:29

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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

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]

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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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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]

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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?

Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink]

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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.

Re: Six distinct pair of shoes are in a closet. If three shoes are [#permalink]

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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