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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
gr gb gy rb ry rg br by bg yr yg yb => 10/12 or 5/6
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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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Re: Four balls in a bag respectively are red, blue, yellow, and [#permalink]
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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