In a drawer of shirts, 8 are blue, 6 are green, and 4 are : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 06:00

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

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

Show Tags

17 Jan 2005, 11:12
00:00

Difficulty:

(N/A)

Question Stats:

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

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: 612
Location: PA
Followers: 5

Kudos [?]: 708 [0], given: 22

Show Tags

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
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

Show Tags

17 Jan 2005, 11:26
There was typo in choice A. I have corrected it.
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 5

Kudos [?]: 708 [0], given: 22

Show Tags

17 Jan 2005, 11:31
ok thanks i was wondering what i did wrong!!
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

Show Tags

17 Jan 2005, 11:33
But the OA is not A...
Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

Show Tags

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
VP
Joined: 18 Nov 2004
Posts: 1440
Followers: 2

Kudos [?]: 37 [0], given: 0

Show Tags

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

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

Show Tags

17 Jan 2005, 14:40
my stupid mistake again
17 Jan 2005, 14:40
Display posts from previous: Sort by