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

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Director
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In a drawer of shirts, 8 are blue, 6 are green, and 4 are [#permalink] New post 17 Jan 2005, 11:12
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 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 that 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

Last edited by gayathri on 17 Jan 2005, 11:25, edited 1 time in total.
Director
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 [#permalink] New post 17 Jan 2005, 11:24
At least one Blue in two draws would be

BB + B not blue + not blue B

8/18 * 7/17 + 8/18 * 10/17 + 10/18 * 8/17

= 25/153

I am not sure what i am doign wrong
Director
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 [#permalink] New post 17 Jan 2005, 11:26
There was typo in choice A. I have corrected it.
Director
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 [#permalink] New post 17 Jan 2005, 11:31
ok thanks i was wondering what i did wrong!! :roll:
Director
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 [#permalink] New post 17 Jan 2005, 11:33
But the OA is not A...
Director
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 [#permalink] New post 17 Jan 2005, 13:49
waysofar wrote:
is the OA 5/17....
i got (10c2)/(18c2)


OA is E

You almost got it. You found the probability of not picking any blue shirt. So picking atleast 1 blue shirt will be 1-5/17 = 12/17
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 [#permalink] New post 17 Jan 2005, 13:52
E. 12/17

Total ways = 18C2

Ways to get atleast 2 blue = 18C2 - 10C2 (ways to get no blue)

p(e) = 12x9 / 9x17 = 12/17
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 [#permalink] New post 17 Jan 2005, 14:40
my stupid mistake again :oops:
  [#permalink] 17 Jan 2005, 14:40
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