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# Four balls in a bag respectively are red, blue, yellow, and

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Senior Manager
Joined: 10 Dec 2004
Posts: 273
Four balls in a bag respectively are red, blue, yellow, and [#permalink]

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22 May 2005, 11:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Four balls in a bag respectively are red, blue, yellow, and green. If two balls are selected at random, what is the probability that one of them is green or blue?
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

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22 May 2005, 12:09
ANS : 1/2

Let the non blue or green colors be rep by X the various possbilities are

XB + BX + XG + GX + BG + GB

in terms of probability it is

6 * ( 1/4 * 1 / 3 ) = 1/2
Director
Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

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22 May 2005, 15:08
5/6

1- prob(red&yellow) = 1 - 1/4c2 = 5/6
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

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22 May 2005, 15:29
gr gb gy rb ry rg br by bg yr yg yb => 10/12 or 5/6
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If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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22 May 2005, 19:30
Also found 5/6

total possible couples : (4*3)/2 = 6
couples of balls with G : GY, GB, GR
couples of balls with B (other than GB) : BY, BR
total positive outcomes (couples G + couples B) / total possible couples
5/6
Manager
Joined: 08 Mar 2005
Posts: 99

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25 May 2005, 00:03
I assumed the ques is at least 1 blue or 1 green.

With that I got 5/6

1 - 1/4 * 1/3 * 2 = 1 - 1/6 = 5/6
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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25 May 2005, 04:20
Agree with 1/2 .We have 4 balls, the prob of taking B OR G is 1/4+1/4=1/2
Manager
Joined: 07 Mar 2005
Posts: 93

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29 May 2005, 13:07
1 - prob(RedvYellow)

RY = 1/12
YR = 1/12

1/12 + 1/12 = 1/6

1- (1/6) = 5/6
SVP
Joined: 05 Apr 2005
Posts: 1710

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30 May 2005, 11:13
4c2=(4x3x2)/(2x2) = 6
green= 1 (3c2)= 3
blue= 1 (3c2)= 3
one of them is green or blue = 3+3-1(common in both)=5
the prob = 5/6
Senior Manager
Joined: 30 Dec 2004
Posts: 294
Location: California

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30 May 2005, 12:04
questions asks green or blue not green and blue therefore:

Probablility blue = 1/4

Probability green = 1/4

Prob green or blue = 1/4 +1/4 = 1/2

Don't see where 5/6 comes from...But then again maybe I'm way off here...
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SVP
Joined: 03 Jan 2005
Posts: 2233

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30 May 2005, 12:19
You are taking two balls, not one ball, greenandwise. That's why you can just add the two 1/4 together.

Basically this question is asking what is the probability that you will get no Red and no White when you draw two balls from the four.

So it would be 1-1/C(4,2)=5/6.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC

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30 May 2005, 16:22
Total outcomes = 4C2 = 6

P(G OR B ) = P(G) + P(B) - P(G AND B )

P(G) = 3/6
P(B) = 3/6
P(G AND B) = 1/6

= 3/6 +3/6 - 1/6
= 5/6
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ash
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I'm crossing the bridge.........

Intern
Joined: 09 Apr 2005
Posts: 5

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31 May 2005, 22:28
P(G or B) first time= 2/4
+
P(G or B) second time=1/3

So P=5/6
Senior Manager
Joined: 17 May 2005
Posts: 271
Location: Auckland, New Zealand

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03 Jun 2005, 13:29
pb_india wrote:
Four balls in a bag respectively are red, blue, yellow, and green. If two balls are selected at random, what is the probability that one of them is green or blue?

when the question states that one of them is green or blue does that mean only one or at least one

i assumed it means only one and got an answer of 2/3

however others have assumed at least one, thus including the possibility of getting green and blue and getting an answer of 5/6

thnx
SVP
Joined: 03 Jan 2005
Posts: 2233

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03 Jun 2005, 21:44
If it says one of two is such and such then the other one could be anything. If it is "exactly one" then the question must have the word "exactly" in there.
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

03 Jun 2005, 21:44
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