ds:probability

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 50
Joined: Wed Apr 06, 2011 11:50 pm

ds:probability

by arjunshn » Wed Apr 13, 2011 4:08 am
A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22

User avatar
Legendary Member
Posts: 1101
Joined: Fri Jan 28, 2011 7:26 am
Thanked: 47 times
Followed by:13 members
GMAT Score:640

by HSPA » Wed Apr 13, 2011 4:16 am
using 1 : 4/12 * 3/11 will do => 4 red

x, 12-x
(x/12 + 12-x-1/11) + 8/12 * 4/11.... got to go havent calculated this..

I guess D
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.

Senior | Next Rank: 100 Posts
Posts: 50
Joined: Wed Apr 06, 2011 11:50 pm

by arjunshn » Wed Apr 13, 2011 4:22 am
but hte answer is a, even i selected d

User avatar
Legendary Member
Posts: 979
Joined: Tue Apr 14, 2009 1:38 am
Location: Hyderabad, India
Thanked: 49 times
Followed by:12 members
GMAT Score:700

by bubbliiiiiiii » Wed Apr 13, 2011 4:22 am
IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Regards,

Pranay

GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

by Testluv » Thu Apr 14, 2011 11:42 pm
bubbliiiiiiii wrote:IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.

You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.

The correct answer is A.
Kaplan Teacher in Toronto

User avatar
Legendary Member
Posts: 979
Joined: Tue Apr 14, 2009 1:38 am
Location: Hyderabad, India
Thanked: 49 times
Followed by:12 members
GMAT Score:700

by bubbliiiiiiii » Sat Apr 16, 2011 5:25 am
Testluv wrote:
bubbliiiiiiii wrote:IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.

You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.

The correct answer is A.
Thanks testluv a great concept shared. :)
Regards,

Pranay