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A box contains three pairs of blue gloves and two pairs of [#permalink]

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A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

Let's calculate the opposite probability of NOT getting a matched set and subtract this value from 1.

This could happen only if we pick all three same hand BLUE gloves; two same hand BLUE gloves and any green glove; or two same hand GREEN gloves and any BLUE glove

BBB: 6/10*2/9*1/8=1/60 (after we pick a blue glove, 6/10, then there is 2 same hand gloves left out of total 9 gloves - 2/9, and so on); BBG: (6/10*2/9*4/8)*3=12/60, multiplying by 3 as this senario can occur in 3 different ways: BBG, BGB, GBB; GGB: (4/10*1/9*6/8)*3=6/60;

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Bleft, Bright, B Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any blue in 4/8 ways. Probability of getting B pair and another B = (3/10)*(3/9)*(4/8)*3!/2! (You multiply by 3! here to account for the order e.g. Bright, Bleft, Bleft or Bleft, Bright, Bright etc but two gloves will be identical so you divide by 2!)

Gleft, Gright, B Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any Blue in 6/8 ways. Probability of getting G pair and a B = (2/10)*(2/9)*(6/8)*3!

Gleft, Gright, G Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any other G in 2/8 ways. Probability of getting G pair and a G = (2/10)*(2/9)*(2/8)*3!/2!

Adding them all up, you get 41/60.

Note here that we cannot say that let's get Bleft, Bright and then any one of the remaining gloves. We need to take separate cases for the third glove (B or G i.e. first two cases above) because the number of arrangements of Bleft, Bright, G is different from number of arrangements of Bleft, Bright, B as we see above. In one case we multiply by 3! because all 3 gloves are distinct. In the other case, we multiply by 3!/2! because 2 of the gloves are identical. Same logic can be used for the green pair. _________________

Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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21 Feb 2012, 11:30

VeritasPrepKarishma wrote:

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Dear Karishma and Bunuel, Please, tell me in which part I am wrong and why.

I analized it in this way: First we are going to see the probabilities when we have a pair of blue gloves =

P(right or left blue glove)* P(the opposite hand blue glove)*P(a green glove) =

But because there could be different arragements of BBG, we multiply 1/10 by 3!/2! So, we have \(\frac{3}{10}\)

I stop here because I know that my analysis is wong but I don't know exactly why. Why is the combination just between colors is not enough? Why do we have to combine left and right hands too? I think that my approach already has considered right and left hands.

Please your comments.

PS. The approach of Bunuel is awesome, but the other approach gives us a great perspective to grasp a ot of concepts related to combinatronics. _________________

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Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

First we are going to see the probabilities when we have a pair of blue gloves =

P(right or left blue glove)* P(the opposite hand blue glove)*P(a green glove) =

But because there could be different arragements of BBG, we multiply 1/10 by 3!/2! So, we have \(\frac{3}{10}\)

[/quote]

Your approach is correct if you know why you did what you did. My only question is this: Why did you multiply by 3!/2! instead of 3! (different arrangements of BBG) since the two Bs are different?

If you understand that when you say 6/10 * 3/9, you are already counting in all arrangements of Bleft and Bright and now all you need to do is arrange G with respect to the 2 Bs (i.e. multiply by 3 for the 3 spots where we can put G), then absolutely, go ahead. There is nothing wrong. (When you say 6/10, you are counting the possibilities of picking a Bleft or a Bright first and whatever is leftover next, so you have already arranged the different Bs.) _________________

Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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24 Mar 2012, 11:20

VeritasPrepKarishma wrote:

arps wrote:

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Hi Karishma. Sorry for replying to an old post. Please take some time and clarify my doubt.

In the above calculation why should we multiply with 3!, How is order important here? while calculating (3/10)*(3/9)*(4/8) we already considered all the possible cominations. Please help! I am getting confused on this type of questions where order is important. Can you suggest some logic so that I can identify the need of considering the order?

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Hi Karishma. Sorry for replying to an old post. Please take some time and clarify my doubt.

In the above calculation why should we multiply with 3!, How is order important here? while calculating (3/10)*(3/9)*(4/8) we already considered all the possible cominations. Please help! I am getting confused on this type of questions where order is important. Can you suggest some logic so that I can identify the need of considering the order?

Thanks, Premasai

One way in which you can get a pair is: Blue Left, Blue Right and any Green. What is the probability that you will get this combination? You can pick Blue Left (probability 3/10 because of the 10 gloves, only 3 are blue left), then you can pick Blue right (probability 3/9 because of the 9 remaining gloves, only 3 are blue right) and then you can pick any green (probability 4/8 because of the 8 remaining gloves, only 4 are green) Probability of this happening = (3/10)*(3/9)*(4/8)

Or You can pick Blue Right (probability 3/10 because of the 10 gloves, only 3 are blue right), then you can pick Blue left (probability 3/9 because of the 9 remaining gloves, only 3 are blue left) and then you can pick any green (probability 4/8 because of the 8 remaining gloves, only 4 are green) Probability of this happening = (3/10)*(3/9)*(4/8)

Or You can pick Green first (probability 4/10 because of the 10 gloves, only 4 are green), then you can pick Blue right (probability 3/9 because of the 9 remaining gloves, only 3 are blue right) and then you can pick blue left (probability 3/8 because of the 8 remaining gloves, only 3 are blue left) Probability of this happening = (4/10)*(3/9)*(3/8)

How many such cases will there be? 3! because you can arrange all 3 in 3! ways. That is the reason you multiply by 3! When you write just "(3/10)*(3/9)*(4/8)", you are considering only one of the possible 6 cases in which you could get Blue right, Blue left and green.

In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

Since there are 3 blue pairs and 2 green pairs, we have a total of 10 gloves.

I considered two cases: Case 1: we have a blue pair match

This can be done if we have a blue left, a blue right and any other glove. So, Since we have 3 blue lefts and 3 blue rights and a total of 10 gloves,

3C1*3C1*8C1- 3 ways of selecting a blue left, 3 for a blue right and any one glove of the remaining 8=72 ways.

You are double counting here. Say the gloves are all distinct. The 3 blue left ones are Bl1, Bl2 and Bl3. Three blue right ones are Br1, Br2, Br3. So you select one of the blue left ones and one of the blue right ones: Bl2, Br3. Now you have 8 leftover and you can select any one of them. Say you select Bl1. So your selection consists of Bl1, Bl2, Br3

Imagine another scenario: So you select one of the blue left ones and one of the blue right ones: Bl1, Br3. Now you have 8 leftover and you can select any one of them. Say you select Bl2. So your selection consists of Bl1, Bl2, Br3

The two selections are the same but you have counted them as different selections.

12bhang wrote:

Case 2: we have a green pair match,

SO, Gleft,Gright and any other glove,

2C1*2C1*8C1=32

Summing , we get 104

The total number of ways to select 3 gloves =10C3=120

so probability of getting a match=104/120 = 13/15.

Where am i going wrong?

Same problem with the green pair. From the solutions given above, review how to effectively use probability to solve this question.

In case you want to use combinations, you still have to take cases:

All three Blues: 3C2*3C1*2 = 18(Select 2 of the blue left and one of the blue right. Multiply by 2 because you can select 2 of the blue right and one of blue left too)

Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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14 Nov 2014, 04:06

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Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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18 Nov 2014, 13:34

Thanks Karishima for your beautiful explanation.

What troubles me is that my instinctive response was not to go through each individual scenario. Instead, I said there are three ways to select a left-handed blue glove, three ways to select a right-handed blue glove and eight ways to select a third glove. Then I said there are two ways to select a left-handed green glove, two ways to select a right-handed green glove and eight ways to select a third glove. I added the two and I got a probability of 13/15, which was clearly much too high. Given my starting point, which as I say was instinctive, is there any way I could have got to the right answer, maybe via a Venn diagram?

What troubles me is that my instinctive response was not to go through each individual scenario. Instead, I said there are three ways to select a left-handed blue glove, three ways to select a right-handed blue glove and eight ways to select a third glove. Then I said there are two ways to select a left-handed green glove, two ways to select a right-handed green glove and eight ways to select a third glove. I added the two and I got a probability of 13/15, which was clearly much too high. Given my starting point, which as I say was instinctive, is there any way I could have got to the right answer, maybe via a Venn diagram?

Note that the left hand blue gloves are all same. This is implied from the statement: "what is the probability that a matched set (i.e., a left- and right-hand glove of the same color)"... So you do not select the left hand blue glove in 3 ways. Instead, the probability of selecting the left hand blue glove is 3/10. Similarly, the logic of selecting right hand blue glove and green gloves is also flawed. _________________

Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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30 Apr 2015, 13:30

VeritasPrepKarishma wrote:

arps wrote:

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Bleft, Bright, B Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any blue in 4/8 ways. Probability of getting B pair and another B = (3/10)*(3/9)*(4/8)*3!/2! (You multiply by 3! here to account for the order e.g. Bright, Bleft, Bleft or Bleft, Bright, Bright etc but two gloves will be identical so you divide by 2!)

Gleft, Gright, B Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any Blue in 6/8 ways. Probability of getting G pair and a B = (2/10)*(2/9)*(6/8)*3!

Gleft, Gright, G Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any other G in 2/8 ways. Probability of getting G pair and a G = (2/10)*(2/9)*(2/8)*3!/2!

Adding them all up, you get 41/60.

Note here that we cannot say that let's get Bleft, Bright and then any one of the remaining gloves. We need to take separate cases for the third glove (B or G i.e. first two cases above) because the number of arrangements of Bleft, Bright, G is different from number of arrangements of Bleft, Bright, B as we see above. In one case we multiply by 3! because all 3 gloves are distinct. In the other case, we multiply by 3!/2! because 2 of the gloves are identical. Same logic can be used for the green pair.

Hi Karishma

In case of BR, BL, B and GL, GR , R. Please explain me why are we dividing by 2.

You have posted a very good explanation but this single thing is bothering me.

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways. Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3! (You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Bleft, Bright, B Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any blue in 4/8 ways. Probability of getting B pair and another B = (3/10)*(3/9)*(4/8)*3!/2! (You multiply by 3! here to account for the order e.g. Bright, Bleft, Bleft or Bleft, Bright, Bright etc but two gloves will be identical so you divide by 2!)

Gleft, Gright, B Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any Blue in 6/8 ways. Probability of getting G pair and a B = (2/10)*(2/9)*(6/8)*3!

Gleft, Gright, G Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any other G in 2/8 ways. Probability of getting G pair and a G = (2/10)*(2/9)*(2/8)*3!/2!

Adding them all up, you get 41/60.

Note here that we cannot say that let's get Bleft, Bright and then any one of the remaining gloves. We need to take separate cases for the third glove (B or G i.e. first two cases above) because the number of arrangements of Bleft, Bright, G is different from number of arrangements of Bleft, Bright, B as we see above. In one case we multiply by 3! because all 3 gloves are distinct. In the other case, we multiply by 3!/2! because 2 of the gloves are identical. Same logic can be used for the green pair.

Hi Karishma

In case of BR, BL, B and GL, GR , R. Please explain me why are we dividing by 2.

You have posted a very good explanation but this single thing is bothering me.

Can you please clear my doubt.

Thanks

Bleft, Bright, B is a combination of 3 blue gloves. The B glove will be either left or right. So 2 of them will be same and one different - you could have 2 blue left gloves and a blue right glove or 2 blue right gloves and a blue left glove.

Did you understand why we multiply by 3!? It's because you can pick 3 different things in 3! ways = 6 ways (Thing1, Thing2, Thing3 or Thing1, Thing3, Thing2 etc). But in how many different ways can you pick 3 things such that 2 of them are identical? You can do that in 3!/2! ways = 3 ways (because 2 are identical) Bleft, Bleft, Bright Bleft, Bright, Bleft Bright, Bleft, Bleft _________________

A box contains three pairs of blue gloves and two pairs of [#permalink]

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01 May 2015, 02:52

Hi Karishma

In case of BR, BL, B and GL, GR , R. Please explain me why are we dividing by 2.

You have posted a very good explanation but this single thing is bothering me.

Can you please clear my doubt.

Thanks[/quote]

Bleft, Bright, B is a combination of 3 blue gloves. The B glove will be either left or right. So 2 of them will be same and one different - you could have 2 blue left gloves and a blue right glove or 2 blue right gloves and a blue left glove.

Did you understand why we multiply by 3!? It's because you can pick 3 different things in 3! ways = 6 ways (Thing1, Thing2, Thing3 or Thing1, Thing3, Thing2 etc). But in how many different ways can you pick 3 things such that 2 of them are identical? You can do that in 3!/2! ways = 3 ways (because 2 are identical) Bleft, Bleft, Bright Bleft, Bright, Bleft Bright, Bleft, Bleft[/quote]

Hi Karishma

I understand your concept.Thanks for the wonderful explanation.

Can you please help me with one thing more.

I assume that order should not matter in this question so we can proceed with combination. so if we follow the below approach, can you please tell me where i am going wrong.

3C1x3C1x4C1 = 1 left blue + 1 Right Blue and 1 Blue 3C1X3C1X4C1 = 1 left blue + 1 right blue + 1 green 2C1X2C1X6C1 = 1 left green + 1 right green + 1 blue 2C1x2C1x2C1 = 1Left Green + 1 Right Green + 1 Green

I assume that order should not matter in this question so we can proceed with combination. so if we follow the below approach, can you please tell me where i am going wrong.

3C1x3C1x4C1 = 1 left blue + 1 Right Blue and 1 Blue 3C1X3C1X4C1 = 1 left blue + 1 right blue + 1 green 2C1X2C1X6C1 = 1 left green + 1 right green + 1 blue 2C1x2C1x2C1 = 1Left Green + 1 Right Green + 1 Green

I have doubt basically in the colored lines.

Awaiting again your wonderful explanation

Thanks in advance!!

You are committing the same mistake:

3C1x3C1x4C1 = 1 left blue + 1 Right Blue and 1 Blue

2 of the selections will be identical. If you want to select, you need to further split them up as

2 left Blue and 1 right Blue in 3C2 * 3C1 = 9 ways 1 left Blue and 2 right Blue in 3C1 * 3C2 = 9 ways

So you can select a left, a right and one any other blue in 9+9 = 18 ways.

Re: A box contains three pairs of blue gloves and two pairs of [#permalink]

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04 May 2015, 09:40

VeritasPrepKarishma wrote:

Radhika11 wrote:

I assume that order should not matter in this question so we can proceed with combination. so if we follow the below approach, can you please tell me where i am going wrong.

3C1x3C1x4C1 = 1 left blue + 1 Right Blue and 1 Blue 3C1X3C1X4C1 = 1 left blue + 1 right blue + 1 green 2C1X2C1X6C1 = 1 left green + 1 right green + 1 blue 2C1x2C1x2C1 = 1Left Green + 1 Right Green + 1 Green

I have doubt basically in the colored lines.

Awaiting again your wonderful explanation

Thanks in advance!!

You are committing the same mistake:

3C1x3C1x4C1 = 1 left blue + 1 Right Blue and 1 Blue

2 of the selections will be identical. If you want to select, you need to further split them up as

2 left Blue and 1 right Blue in 3C2 * 3C1 = 9 ways 1 left Blue and 2 right Blue in 3C1 * 3C2 = 9 ways

So you can select a left, a right and one any other blue in 9+9 = 18 ways.

Same for green.

Hi Karishma

Thanks Again

If i understood the concept right, then to get atleast 1 get pair from 3 blue socks, we can also do like below 3C1.3C1.2C1-> 1 Left Blue + 1 Right Blue + 1 Blue Because if we select any of the left 2 Blue sock (or 2 right blue sock), we will get the same kind of combination so we can discard one left blue and one right blue sock and will choose among the remaining 2 socks so 2C1.

If i understood the concept right, then to get atleast 1 get pair from 3 blue socks, we can also do like below 3C1.3C1.2C1-> 1 Left Blue + 1 Right Blue + 1 Blue Because if we select any of the left 2 Blue sock (or 2 right blue sock), we will get the same kind of combination so we can discard one left blue and one right blue sock and will choose among the remaining 2 socks so 2C1.

Am i wrong in interpreting the concept ?

But the remaining two socks will also be a left and a right! Every pair of socks will have a sock for the left foot (with space for big toe on the right) and one for the right foot (with space for the big toe on the left). So basically you have 3 identical left blue socks and 3 identical right foot socks. In how many ways can you select 3 socks out of these 6 such that you get a pair to wear? You can select either 2 left and 1 right or 2 right and 1 left. _________________

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