It is currently 11 Dec 2017, 04:37

### 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

# There 4 balls - yellow, red, blue, and green. You pick two

Author Message
Director
Joined: 14 Sep 2005
Posts: 984

Kudos [?]: 226 [0], given: 0

Location: South Korea
There 4 balls - yellow, red, blue, and green. You pick two [#permalink]

### Show Tags

24 Nov 2005, 05:43
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.

There 4 balls - yellow, red, blue, and green. You pick two of them. In how many cases one of the two is yellow or green?
_________________

Auge um Auge, Zahn um Zahn !

Kudos [?]: 226 [0], given: 0

Director
Joined: 24 Oct 2005
Posts: 658

Kudos [?]: 16 [0], given: 0

Location: London

### Show Tags

24 Nov 2005, 08:52
I just realised that is not probability that is asked.
So no of cases I think with yellow or green 10 out of the total 12.

Last edited by remgeo on 24 Nov 2005, 09:19, edited 2 times in total.

Kudos [?]: 16 [0], given: 0

SVP
Joined: 05 Apr 2005
Posts: 1705

Kudos [?]: 101 [0], given: 0

Re: PS - Selecting balls [#permalink]

### Show Tags

24 Nov 2005, 08:58
total = 4c2=6
not either yello or green = RB= 1
so the total possibilities = 6-1= 5, (YG is also considered the possibile case).

Kudos [?]: 101 [0], given: 0

Current Student
Joined: 29 Jan 2005
Posts: 5201

Kudos [?]: 439 [0], given: 0

### Show Tags

24 Nov 2005, 09:02
4P2= 12 possible ways to select the balls (unless they are replaced after removal)

only 2 constraints exist: red/blue or blue/red

12-2= 10 ways

Or in terms of probability: 10/12--->5/6

Kudos [?]: 439 [0], given: 0

24 Nov 2005, 09:02
Display posts from previous: Sort by