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In a drawer of shirts 8 are blue, 6 are green and 4 are
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Updated on: 27 Oct 2013, 06:28
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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? (A) 25/153 (B) 28/153 (C) 5/17 (D) 4/9 (E) 12/17
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Originally posted by Bhai on 11 Dec 2005, 20:07.
Last edited by Bunuel on 27 Oct 2013, 06:28, edited 2 times in total.
Edited the question, added the OA and moved to PS forum.



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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



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Does it matter if he draws the shirts at once or in order.....



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I think its E
there are a total of 18 shirts. 8 blue and 10 non blue.
P(selecting at least 1 blue shirt) = 1  P(selecting no blue shirts)
Assuming no replacement
P(selecting first nonblue shirt) = 10/18
P(selecting secong nonblue shirt) = 9/17
P(selecting no blue shirts) = 10/18 * 9/17= 10/34
Therefor P(selecting at least 1 blue shirt) = 1 (10/34)= 24/34=12/17



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Lucky Twin + intuition guess (must be more than 50%) 12/17.
(E)



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Re: Probability:Blue Shirt
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14 Dec 2005, 01:16
Bhai wrote: 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



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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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26 Oct 2013, 15:51
Its may be blue but i m not sure....



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ywilfred wrote: 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 . 80/153 + 28/153 = 12/17



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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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27 Oct 2013, 06:32



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TeHCM wrote: Does it matter if he draws the shirts at once or in order..... Mathematically the probability of picking 2 shirts simultaneously, or picking them one at a time (without replacement) is the same.
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Hi Bunuel, Please find my doubt below.
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



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kona wrote: Hi Bunuel, Please find my doubt below.
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 notblue shirts out of 10. 2. No. You can pick randomly onebyone as well as simultaneously. 3. Hope this helps.
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kona wrote: Hi Bunuel, Please find my doubt below.
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) tshirts. 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) Please press Kudos if you wish to appreciate*** Kudos is the best form of appreciation***



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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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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 tileowners 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 newborn 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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kona wrote: 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?
Check out this post for a discussion on when order matters and when it doesn't: http://www.veritasprep.com/blog/2013/08 ... ermatter/
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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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22 May 2018, 18:09
Bhai wrote: 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 We can use the equation: P(at least one of the shirts he draws will be blue) = 1  P(no blue shirts) P(no blue shirts) = 10/18 x 9/17 = 5/9 x 9/17 = 5/17. So, P(at least one of the shirts he draws will be blue) = 1  5/17 = 12/17. Answer: E
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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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22 May 2018, 22:11
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?
Best way to calculate is to find the probability of drawing no blue shirts in both the attempts.
Probability of selecting at least 1 blue shirt = 1  p(no blue shirt)
Probability of not selecting blue shirt in first draw = \(\frac{10}{18}\) = \(\frac{5}{9}\)
Probability of not selecting blue shirt in second draw = \(\frac{9}{17}\)
Probability of not selecting blue shirt = \(\frac{5}{9}\)*\(\frac{9}{17}\) = \(\frac{5}{17}\)
Probability of selecting at least 1 blue shirt = 1  \(\frac{5}{17}\) = \(\frac{12}{17}\)
Ans: D



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Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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24 Nov 2019, 03:59
Bhai wrote: 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 total = 18 shirts and to get atleast 1 blue = (1 (10/18*9/17) ) => 12/17 IMO E




Re: In a drawer of shirts 8 are blue, 6 are green and 4 are
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24 Nov 2019, 03:59






