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Sub 505 (Easy)|   Probability|               
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Lolaergasheva
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 08 Mar 2011, 09:45
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E
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5% (low)

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90% (00:57) correct 10% (00:57) wrong based on 1505 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


(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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E
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beyondgmatscore
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 08 Mar 2011, 10:19
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Lolaergasheva
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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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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siddhans
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 30 Sep 2011, 17:02
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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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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 30 Sep 2011, 18:41
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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
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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siddhans
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 30 Sep 2011, 19:19
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Spidy001
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
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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If there are 5 red eggs doesnt probability of selecting one red ball out of 5 be 1/5?
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 30 Sep 2011, 19:41
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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
Spidy001
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
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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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 26 Jul 2019, 05:51
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Lolaergasheva
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
Show more

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

Answer: E

Cheers,
Brent
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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 28 Jul 2020, 06:41
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Lolaergasheva
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
Show more

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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Each of the eggs in a bowl is dyed red, or green, or blue. If one egg 04 Oct 2025, 07:15
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