Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]
02 Sep 2006, 16:00

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

56% (02:12) correct
44% (01:08) wrong based on 225 sessions

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.

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.

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)

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?

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.

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.

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 _________________

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

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

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. _________________

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......

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.

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. _________________

Re: DS-700 level-Probability [#permalink]
05 Aug 2011, 13:25

(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).

Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]
16 Dec 2011, 20:47

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. _________________

Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]
17 Nov 2014, 10:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]
20 Nov 2014, 04:42

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.

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.

Re: A box of doughnuts contains glazed, cream-filled, and jelly [#permalink]
20 Nov 2014, 07:43

Expert's post

2

This post was 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.

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.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Last week, hundreds of first-year and second-year students traversed the globe as part of KWEST: Kellogg Worldwide Experience and Service Trip. Kyle Burr, one of the student-run KWEST executive...