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In a drawer of shirts 8 are blue, 6 are green and 4 are [#permalink]

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11 Dec 2005, 20:07

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In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

16. In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A)25/153 (B)28/153 (C)5/17 (D)4/9 (E)12/17

= 1 - p(None of them being blue)
= 1 - (10/18 * 9 /17)
= 1 - 5/17
= 12/17 E

Assume no replacement: P(1 blue and 1 other colour) = 8/18 * 10/17 = 40/153 P(1 blue and 1 blue) = 8/18 * 7/17 = 28/153

40/153 + 28/153 = 4/9 is the probability at least one is blue

Hey, I think you went wrong here because there are 2 possibilities of him picking one blue and one other color.. Blue first then Random or Random first then Blue.. Which will double the probability .

In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A) 25/153 (B) 28/153 (C) 5/17 (D) 4/9 (E) 12/17

P(at least one blue) = 1- P(0 blue) = 1 - 10/18*9/17 = 12/17.

q:In a drawer of shirts, 8 are blue, 6 are green, and 4 are red. If Miles draws two shirts at random, what is the probability that at least one of the shirts he draws is blue?

My approach:1-(zero shirts blue)

1-(10*9)/18c2

1-(10/17)=7/17.

Wrong..

Another approach:1-(10/18*9/17)

1-(5/17)=12/17

Right..

My doubts:1.When picking two shirts at random from a group of 18,is it not 18c2 for a total number of possibilities

2.Also,does "picking at random" means "picking simultaneously?"

3.Does ordering in this case matter?When should the order be considered and when it shouldnt be?Could you please throw some light on this?

q:In a drawer of shirts, 8 are blue, 6 are green, and 4 are red. If Miles draws two shirts at random, what is the probability that at least one of the shirts he draws is blue?

My approach:1-(zero shirts blue)

1-(10*9)/18c2

1-(10/17)=7/17.

Wrong..

Another approach:1-(10/18*9/17)

1-(5/17)=12/17

Right..

My doubts:1.When picking two shirts at random from a group of 18,is it not 18c2 for a total number of possibilities

2.Also,does "picking at random" means "picking simultaneously?"

3.Does ordering in this case matter?When should the order be considered and when it shouldnt be?Could you please throw some light on this?

Please rectify

Thank you

Regards, Kona

1. If you use combination approach then it should be \(P(at \ least \ one \ blue) = 1- P(0 \ blue) = 1 - \frac{C^2_{10}}{C^2_{18}} = \frac{12}{17}\). So, your denominator was correct but the numerator was not. It should have been \(C^2_{10}\): the number of ways to pick two not-blue shirts out of 10.

2. No. You can pick randomly one-by-one as well as simultaneously.

q:In a drawer of shirts, 8 are blue, 6 are green, and 4 are red. If Miles draws two shirts at random, what is the probability that at least one of the shirts he draws is blue?

My approach:1-(zero shirts blue)

1-(10*9)/18c2

1-(10/17)=7/17.

Wrong..

Another approach:1-(10/18*9/17)

1-(5/17)=12/17

Right..

My doubts:1.When picking two shirts at random from a group of 18,is it not 18c2 for a total number of possibilities

2.Also,does "picking at random" means "picking simultaneously?"

3.Does ordering in this case matter?When should the order be considered and when it shouldnt be?Could you please throw some light on this?

Please rectify

Thank you

Regards, Kona

Hi Kona, in your 1st approach , it should be : 1- (10C2)/(18C2) = 1 -5/17 = 12/17

Here 10C2 defines the no. of ways in which 2 shirts can be picked from 10 (6G +4R) t-shirts. Your numerator calcn 10 *9 makes a difference b/w a particular pair of 1G1R and 1R1G, which are essentially the same. So,for your calcn of numerator double counting is happening.

1.Denominator 18C2 is perfectly ok for picking two shirts at random/simultaneously 2.It means the same in the present question context 3.Order would not matter as it is the case of picking (and not arranging thereafter)

Re: In a drawer of shirts 8 are blue, 6 are green and 4 are [#permalink]

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22 May 2014, 14:30

Dear Kona

You asked: "When should the order be considered and when it shouldn't be?"

I'll try to help you with this doubt.

Let's say there are 26 small tiles lying in a heap. Each tile has one letter of the English alphabet inscribed on it, and no two tiles have the same alphabet. Now you are asked by the tile-owners to pick up 2 tiles from this heap.

They are simply asking you to pick up the tiles; they are not interested in the order in which you pick up the tiles. Nothing will happen to one tile because it was picked up first, or because it was picked up second. So, whether you pick up E first and J later, or J first and E later doesn't matter. So, you will simply apply the formula to select a bunch of r things out of a bunch of n things =nCr.

When would order matter?

It would matter if they had told you that they will start the name of their new-born baby with the alphabet on the first tile you picked up. And that if the alphabet on your second slide lies between A and G, you will get $10. So, in this case, there are consequences to the order in which the tiles were picked up. You may still have picked up E and J, but the order in which you picked up will decide whether the baby will be named Eugene or Jeannie, and whether you will get $10 or not.

So, let's now tackle a question: What is the probability that the baby's name started with E and you got $10?

Total number of ways to pick up 1st tile = 26 Total number of ways to pick up 2nd tile = 25 So, Total number of ways to pick up the two tiles = 26*25

(Note that we didn't take 26C2, because 26C2 applies when you are taking out a bunch of 2 things out of a bunch of 26 things and order doesn't matter)

Favorable Cases: Total number of ways in which the 1st tile can be E = 1 Total number of ways in which $10 can be won on the second tile = 6 (There are 7 alphabets from A to G, but E has already been taken out in the first pick)

So, Desired probability = (1*6)/(26*25) =3/325
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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are [#permalink]

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23 Aug 2015, 21:50

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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are [#permalink]

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26 Apr 2017, 10:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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