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

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

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22 May 2005, 10:25
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?
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22 May 2005, 11:09
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
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22 May 2005, 14:08
5/6

1- prob(red&yellow) = 1 - 1/4c2 = 5/6
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22 May 2005, 14:29
gr gb gy rb ry rg br by bg yr yg yb => 10/12 or 5/6
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22 May 2005, 18:30
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
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24 May 2005, 23:03
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
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25 May 2005, 03:20
Agree with 1/2 .We have 4 balls, the prob of taking B OR G is 1/4+1/4=1/2
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29 May 2005, 12:07
1 - prob(RedvYellow)

RY = 1/12
YR = 1/12

1/12 + 1/12 = 1/6

1- (1/6) = 5/6
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Re: Probab - balls [#permalink]

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30 May 2005, 10:13
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
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30 May 2005, 11:04
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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30 May 2005, 11:19
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.
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30 May 2005, 15:22
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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31 May 2005, 21:28
P(G or B) first time= 2/4
+
P(G or B) second time=1/3

So P=5/6
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Re: Probab - balls [#permalink]

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03 Jun 2005, 12:29
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

please clarify about how to make a distiction

thnx
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03 Jun 2005, 20:44
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;
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03 Jun 2005, 20:44
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