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# A box of doughnuts contains glazed, cream-filled, and jelly

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A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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02 Sep 2006, 17:00
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A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.

[Reveal] Spoiler: Doubt
Why is 1) Not sufficient? If P of second donut selected = 2/5 isn't P of first donut 2+1 / 5+1 = 3/6? (i.e. we put donut back in the box).

Here is what Kaplan said, but I don't get why 1) is not suff.

1)+2):
If there were originally 5 glazed doughnuts in the box, and one was
selected, then there are 4 glazed doughnuts in the box. Since the
probability of picking another glazed doughnut is 2/5, you know that
2/5 = 4 / num of d left, so
there must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Thank you kindly.
[Reveal] Spoiler: OA
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02 Sep 2006, 18:18
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Is the answer C i.e. both statements are needed ?

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Hence INSUFF

Statement 2 says there were 5 glazed doughnuts originally
NOT SUFF

Combining, let x be the total no. of doughnuts, then

2/5 x = 4 (since we know that there were 5 originally)

i.e. x = 20/2 = 10

Hence required prob. = 5/10 = 0.5
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02 Sep 2006, 18:58
gmatornot wrote:
Is the answer C i.e. both statements are needed ?

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Hence INSUFF

Statement 2 says there were 5 glazed doughnuts originally
NOT SUFF

Combining, let x be the total no. of doughnuts, then

2/5 x = 4 (since we know that there were 5 originally)

i.e. x = 20/2 = 10

Hence required prob. = 5/10 = 0.5

Correct the answer is C and your explanation matches except you forgot to add 1 to 10 making it P= 5/11.

But I still don't understand. If after removing 1st donut P=2/5, if we want to go back 1 step can't we just add 1 to both num and denom making it 3/6 =1/2?
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02 Sep 2006, 19:29

Nsentra,

For the 1 statement we cannot add 1 to the den and num coz we don't know what the fraction would be it can be 2/5 or 4/10 or 6/15 which will give different answers every time.
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02 Sep 2006, 19:44
smily_buddy wrote:

Nsentra,

For the 1 statement we cannot add 1 to the den and num coz we don't know what the fraction would be it can be 2/5 or 4/10 or 6/15 which will give different answers every time.

Ah, ok I got it. I see, this goes in hand with a type of question that asks if it's possible to find a new ratio when something was added/subtracted to some part of current ratio.

Thank you!
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16 Nov 2008, 11:28
C.

Stmt2 gives number of glazed doughnuts and stmt1 gives total number of doughnuts.
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16 Nov 2008, 14:44
scthakur wrote:
C.

Stmt2 gives number of glazed doughnuts and stmt1 gives total number of doughnuts.

I am not sure how stmt1 is giving the total number of DN's.

The probability of selecting a second glazed DN is 2/5. We need P of selecting the first DN.

Are you saying that the total number of DN's is 10 because 4/10 is 2/5?

I picked E because the problem does not tell us whether the first DN is put back or not. I see where you are getting at but some how I cannot explain
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16 Nov 2008, 14:48
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5
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16 Nov 2008, 14:53
GMAT TIGER wrote:
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5

I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5
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17 Nov 2008, 00:33
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study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

(1)
say there are x glazed donuts, and "n" total donuts
for the first glazed donut prob = x/n
prob for second = x-1/n-1 = 2/5
But as this is ratio we can not conclude anything about x and n (it can be anythign 2/5, 10/25 etc)
insuff

(2)
we know that x = 5 but insuff. alone

together,
x-1/n-1=2/5 and x = 5
we get, 10=n-1 or n = 11

Suff to calculate prob the doughnut selected is glazed

C
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17 Nov 2008, 15:07
icandy wrote:
GMAT TIGER wrote:
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5

I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.
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19 Nov 2008, 02:17
GMAT TIGER wrote:

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.

IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......
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19 Nov 2008, 03:19
alpha_plus_gamma wrote:
GMAT TIGER wrote:

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.

IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......

Actually, the question would have speicified that the second probability is after the replacement, then there was no value in the question. But again.. I think it is responsibility of the question maker to avoid any ambiguity to us.
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06 Sep 2009, 02:14
IMO C too. Agree with you, guys.

I think that first statement gives the probability of selecting second doughnut without putting it back. Thus, together with statement 2 it gives us the number of total doughnuts and the number of glazed doughnuts.
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05 Aug 2011, 14:25
(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

This does NOT mean that the first chance is 3/6 = 1/2 because if there were, for example, 4 glazed and 6 other donuts after the first draw (4/10 = 2/5 chance), then before that there were 5 glazed and 6 others (5/11 chance). Insufficient.

(2) There were originally 5 glazed doughnuts in the box.

Obviously insufficient on its own since we need to know the total number BUT, with the first one it is sufficient:

If there were 5 glazed and one was drawn, there are now 4 glazed left. There's a 2/5 chance to draw another glazed so there are 4 glazed and 6 others (10 total). So before there were 5 glazed and 6 others (5/11 chance).

C
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Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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16 Dec 2011, 21:47
Let 'g' be the number of glazed doughnuts and let 'n' be the total number of doughnuts.
Thus, we need to find the value of (g / n).

Statement 1:
After the first glazed doughnut is selected, the number of remaining glazed doughnuts = (g - 1)
Number of remaining total doughnuts = (n - 1)
Thus this statement tells us that (g - 1)/(n - 1) = 2/5 => (5g - 5) = (2n - 2) => 5g = 2n + 3
As there are two unknowns, we cannot get a value for (g/n). INSUFFICIENT.

Statement 2: g = 5
This does not give a value for n. So INSUFFICIENT.

Combining these two statements, we have the value of 'g' from (2), which can be substituted in the equation obtained from (1) to find the value of 'n'.
Thus, we can find out (g/n). SUFFICIENT.
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Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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20 Nov 2014, 05:42
Nsentra wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.

[Reveal] Spoiler: Doubt
Why is 1) Not sufficient? If P of second donut selected = 2/5 isn't P of first donut 2+1 / 5+1 = 3/6? (i.e. we put donut back in the box).

Here is what Kaplan said, but I don't get why 1) is not suff.

1)+2):
If there were originally 5 glazed doughnuts in the box, and one was
selected, then there are 4 glazed doughnuts in the box. Since the
probability of picking another glazed doughnut is 2/5, you know that
2/5 = 4 / num of d left, so
there must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Thank you kindly.

Experts,

Could you pls help to solve this DS.

Regards,
Ammu
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Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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20 Nov 2014, 08:43
Expert's post
2
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BOOKMARKED
ammuseeru wrote:
Nsentra wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.

[Reveal] Spoiler: Doubt
Why is 1) Not sufficient? If P of second donut selected = 2/5 isn't P of first donut 2+1 / 5+1 = 3/6? (i.e. we put donut back in the box).

Here is what Kaplan said, but I don't get why 1) is not suff.

1)+2):
If there were originally 5 glazed doughnuts in the box, and one was
selected, then there are 4 glazed doughnuts in the box. Since the
probability of picking another glazed doughnut is 2/5, you know that
2/5 = 4 / num of d left, so
there must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Thank you kindly.

Experts,

Could you pls help to solve this DS.

Regards,
Ammu

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.

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Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]

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31 Mar 2016, 18:28
Nsentra wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.

pretty straight forward question..looks like a 600-700 than a 700 lvl...

we need to find: G/(G+C+J)

1. G-1/(G+C+J-1) = 2/5
it could have been 3 G, and probability is 1/2
or it could be 5 G, and 11 total, thus probability is 5/11.
1 alone is insufficient.

2 alone is insufficient.

1+2
we know that originally, there were 5 G.
so 2x/5x => 2x=5-1 = 4, so x=2. or total there were 11 donuts, out of which 5 were G.

sufficient.
C
Re: A box of doughnuts contains glazed, cream-filled, and jelly   [#permalink] 31 Mar 2016, 18:28
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