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.
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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) =

\(\frac{6}{10} * \frac{3}{9} * \frac{4}{8} = \frac{1}{10}\)

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.)
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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.

I have put up some posts on probability here:
http://www.veritasprep.com/blog/categor ... er-wisdom/

They might help you get better clarity.
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Same for me. Why doesn't that work for this question when it works for this for example:

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?

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

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 have doubt basically in the colored lines.

Awaiting again your wonderful explanation

Thanks in advance!!

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.

Am i wrong in interpreting the concept ?

Hi,
Case where BBB is considered probability to pick a blue glove is 6/10.
Now the second blue glove can only be selected from the remaining 4 Blue gloves as the matching air of the first blue glove has to be ignored.

So shouldn't it come out to be 4/9 instead of 2/9