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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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Re: Each of the eggs in a bowl is dyed red, or green, or blue. If one egg [#permalink]
2
Kudos
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).

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]
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).

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]
1
Kudos
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).

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]
1
Kudos
Top Contributor
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

Target question: What is the probability that the egg will be green?
Upon scanning the two statements, it certainly seems like there isn't enough information (each statement mentions just 1 color, and neither statement mentions green eggs)
As such, we can probably head straight to....
.
.
.
Statements 1 and 2 combined
There are several scenarios that satisfy BOTH statements. Here are two:
Case a: There are 5 red eggs, 1 green egg and 3 blue eggs (notice that both statements are satisfied). In this case, the answer to the target question is P(select green egg) = 1/9
Case b: There are 5 red eggs, 3 green egg and 4 blue eggs (notice that both statements are satisfied). In this case, the answer to the target question is P(select green egg) = 3/12 = 1/4
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

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

Probability (green egg drawn) = # favorable events of drawing a green egg / #total no. of elementary events or sample space

P(GE) = n (green eggs)/N (total eggs)

Therefore, to calculate P(GE), we need values for both

'n' = total no. of green eggs and
'N' = total no. of eggs

Statement 1: Only gives us total no. of red eggs. Can't calculate P (GE) using this value.
REJECT A, D

Statement 2: Gives value of P (BE) i.e. probability of a blue egg drawn. This also can't help us in calculating P(GE)
REJECT B

Statement 1 & 2 together: This neither gives us value of 'n' and 'N'
Reject D

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