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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg

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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 08 Mar 2011, 08:45
2
00:00
A
B
C
D
E

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  5% (low)

Question Stats:

83% (00:54) correct 17% (00:36) wrong based on 179 sessions

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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg is to be removed at random. what is the probability that the egg will be green?

(1) There are 5 red eggs in the bowl.
(2) The probability that the egg will be blue is 1/3
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Manager
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 08 Mar 2011, 09:19
5
Lolaergasheva wrote:
Each of the eggs in a bowl is dyed red, or green, or blue. If one egg is to be removed at random. what is the probability that the egg will be green?
(1) There are 5 red eggs in the bowl.
(2) The probability that the egg will be blue is 1/3


If the bowl has R red, G green and B blue eggs, then probability of a green egg would be G/(R+G+B).

Statement 1 says, R=5, but we do not know anything about G and B. So ,insufficient

Statement 2 says, B/(R+G+B) = 1/3, which means that (R+G)/(R+G+B) = 2/3, but we still don't know enough to calculate G/(R+G+B) So , insufficient

Combining 1 and 2, we know (5+G)/(5+G+B) = 2/3, but this still has two variables, so cant solve.

Hence, E
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 30 Sep 2011, 16:02
From st 1 we get : probability of selecting 1 red = 1/5 --- Insufficient

From st 2 we get: probability of selecting 1 blue = 1/3 ---- Insufficient

Combining 1 and 2 we get =>

Probability of 1 red or 1 blue = 1/5 + 1/3

Probability of green = 1 - (1/5 + 1/3)



Hence should be C... What wrong with this logic ...Can someone explain the flaw?
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 30 Sep 2011, 17:41
2
P(R) is not 1/5.

P(R) = R/(total)

Given that there are 5 Red . But we don't know the total . so we cannot calculate P(R) without knowing total.


1. Not sufficient
R=5
when G = 2 , B = 1 => P(G) = 2/8
when G=1, B=2 => P(G) = 1/8

2. Not Sufficient

P(B) = 1/3 = x/3x

But this doesn't tell us anything about B or total because P(B) = 1/3 , but it could also be 2/6 or 3/9 so on..

So we can have different B . Thus we can have different G combinations.

Together - Not Sufficient

P(B) = x/(3x) , R = 5

P(G) = 1- P( not G )

= 1- P(R or B)

But we cannot calculate this with out knowing P(R) and P(B).

Hence answer is E.





siddhans wrote:
From st 1 we get : probability of selecting 1 red = 1/5 --- Insufficient

From st 2 we get: probability of selecting 1 blue = 1/3 ---- Insufficient

Combining 1 and 2 we get =>

Probability of 1 red or 1 blue = 1/5 + 1/3

Probability of green = 1 - (1/5 + 1/3)



Hence should be C... What wrong with this logic ...Can someone explain the flaw?
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 30 Sep 2011, 18:19
Spidy001 wrote:
P(R) is not 1/5.

P(R) = R/(total)

Given that there are 5 Red . But we don't know the total . so we cannot calculate P(R) without knowing total.


1. Not sufficient
R=5
when G = 2 , B = 1 => P(G) = 2/8
when G=1, B=2 => P(G) = 1/8

2. Not Sufficient

P(B) = 1/3 = x/3x

But this doesn't tell us anything about B or total because P(B) = 1/3 , but it could also be 2/6 or 3/9 so on..

So we can have different B . Thus we can have different G combinations.

Together - Not Sufficient

P(B) = x/(3x) , R = 5

P(G) = 1- P( not G )

= 1- P(R or B)

But we cannot calculate this with out knowing P(R) and P(B).

Hence answer is E.





siddhans wrote:
From st 1 we get : probability of selecting 1 red = 1/5 --- Insufficient

From st 2 we get: probability of selecting 1 blue = 1/3 ---- Insufficient

Combining 1 and 2 we get =>

Probability of 1 red or 1 blue = 1/5 + 1/3

Probability of green = 1 - (1/5 + 1/3)



Hence should be C... What wrong with this logic ...Can someone explain the flaw?


If there are 5 red eggs doesnt probability of selecting one red ball out of 5 be 1/5?
Director
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 30 Sep 2011, 18:41
1
No. You are not selecting 1 ball out of 5 red balls.

Here we are talking about probability of with drawing 1 red ball out of n balls out of which lets say R are red.

probability = possible out comes/ total out comes

P(R=1) = Rc1/nc1


if you were to select 1 red ball out of 5 red balls then P(R=1) = 1/5.

siddhans wrote:
Spidy001 wrote:
P(R) is not 1/5.

P(R) = R/(total)

Given that there are 5 Red . But we don't know the total . so we cannot calculate P(R) without knowing total.


1. Not sufficient
R=5
when G = 2 , B = 1 => P(G) = 2/8
when G=1, B=2 => P(G) = 1/8

2. Not Sufficient

P(B) = 1/3 = x/3x

But this doesn't tell us anything about B or total because P(B) = 1/3 , but it could also be 2/6 or 3/9 so on..

So we can have different B . Thus we can have different G combinations.

Together - Not Sufficient

P(B) = x/(3x) , R = 5

P(G) = 1- P( not G )

= 1- P(R or B)

But we cannot calculate this with out knowing P(R) and P(B).

Hence answer is E.





siddhans wrote:
From st 1 we get : probability of selecting 1 red = 1/5 --- Insufficient

From st 2 we get: probability of selecting 1 blue = 1/3 ---- Insufficient

Combining 1 and 2 we get =>

Probability of 1 red or 1 blue = 1/5 + 1/3

Probability of green = 1 - (1/5 + 1/3)



Hence should be C... What wrong with this logic ...Can someone explain the flaw?


If there are 5 red eggs doesnt probability of selecting one red ball out of 5 be 1/5?
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg  [#permalink]

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New post 30 Sep 2018, 10:34
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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg &nbs [#permalink] 30 Sep 2018, 10:34
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