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

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Director
Joined: 07 Nov 2004
Posts: 694
Followers: 4

Kudos [?]: 29 [0], given: 0

In a drawer of shirts, 8 are blue, 6 are green, and 4 are [#permalink]  17 Jan 2005, 11:12
00:00

Difficulty:

(N/A)

Question Stats:

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
Joined: 07 Jun 2004
Posts: 614
Location: PA
Followers: 3

Kudos [?]: 276 [0], given: 22

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
Joined: 07 Nov 2004
Posts: 694
Followers: 4

Kudos [?]: 29 [0], given: 0

There was typo in choice A. I have corrected it.
Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
Followers: 3

Kudos [?]: 276 [0], given: 22

ok thanks i was wondering what i did wrong!!
Director
Joined: 07 Nov 2004
Posts: 694
Followers: 4

Kudos [?]: 29 [0], given: 0

But the OA is not A...
Director
Joined: 07 Nov 2004
Posts: 694
Followers: 4

Kudos [?]: 29 [0], given: 0

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
VP
Joined: 18 Nov 2004
Posts: 1442
Followers: 2

Kudos [?]: 20 [0], given: 0

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
Intern
Joined: 04 Jan 2005
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

my stupid mistake again
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