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In a box there are A green balls, 3A + 6 red balls and 2

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SVP
SVP
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Joined: 05 Apr 2005
Posts: 1732
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In a box there are A green balls, 3A + 6 red balls and 2 [#permalink] New post 14 Aug 2005, 19:42
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (01:47) correct 0% (00:00) wrong based on 2 sessions
In a box there are A green balls, 3A + 6 red balls and 2 yellow ones. If there are no other colors, what is the probability of taking out a green or a yellow ball?

1/5.
½.
1/3.
¼.
2/3.
Intern
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Joined: 13 Aug 2005
Posts: 26
Location: Israel
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Kudos [?]: 2 [0], given: 0

 [#permalink] New post 14 Aug 2005, 22:41
Total balls in the box: 4a+8
The probability of taking a green: a/(4a+8)
The probability of taking a yellow: 2/(4a+8)
The probability of taking G or Y: a/(4a+8)+ 2/(4a+8) =
= a/4(a+2)+ 2/4(a+2) =
= (a+2)/4(a+2) = 1/4
-> Answer D
Senior Manager
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Re: PS: Balls [#permalink] New post 15 Aug 2005, 08:28
HIMALAYA wrote:
In a box there are A green balls, 3A + 6 red balls and 2 yellow ones. If there are no other colors, what is the probability of taking out a green or a yellow ball?

1/5.
½.
1/3.
¼.
2/3.


another D 1/4
so i pick a value for A
let A be 10 GREEN 1O
So red be 36
and yellow 2
TOTAL 48

PROB TO HAVE GREEN 10/48
PROB TO HAVE YELLOW 2/48

Or mean addition principle
so 12/48
OR 1/4
Senior Manager
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Posts: 374
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Re: PS: Balls [#permalink] New post 17 Aug 2005, 01:48
HIMALAYA wrote:
In a box there are A green balls, 3A + 6 red balls and 2 yellow ones. If there are no other colors, what is the probability of taking out a green or a yellow ball?

1/5.
½.
1/3.
¼.
2/3.


P(G or Y) = 1 - P(R) = 1 - (3A+6)/(4A+8) = 1 - 3/4 = 1/4
Re: PS: Balls   [#permalink] 17 Aug 2005, 01:48
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In a box there are A green balls, 3A + 6 red balls and 2

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