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In a drawer of shirts 8 are blue, 6 are green and 4 are

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In a drawer of shirts 8 are blue, 6 are green and 4 are [#permalink] New post 11 Dec 2005, 19:07
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59% (02:39) correct 40% (01:24) wrong based on 47 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Oct 2013, 05:28, edited 2 times in total.
Edited the question, added the OA and moved to PS forum.
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 [#permalink] New post 11 Dec 2005, 19:39
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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 [#permalink] New post 11 Dec 2005, 19:50
Does it matter if he draws the shirts at once or in order.....
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 [#permalink] New post 12 Dec 2005, 00:45
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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 non-blue shirt) = 10/18
P(selecting secong non-blue 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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 [#permalink] New post 12 Dec 2005, 07:07
Lucky Twin + intuition guess (must be more than 50%) 12/17.

(E)
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Re: Probability:Blue Shirt [#permalink] New post 14 Dec 2005, 00:16
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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 [#permalink] New post 26 Oct 2013, 14:51
Its may be blue but i m not sure....
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Re: [#permalink] New post 26 Oct 2013, 20:58
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 [#permalink] New post 27 Oct 2013, 05:32
Expert's post
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


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

Answer: E.

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Re: [#permalink] New post 27 Oct 2013, 05:34
Expert's post
Re:   [#permalink] 27 Oct 2013, 05:34
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