Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 02:24

### GMAT Club Daily Prep

#### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A box contains 100 tickets marked 1 to 100. One ticket is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 23 Apr 2010
Posts: 581
Followers: 2

Kudos [?]: 86 [3] , given: 7

A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

07 Feb 2011, 02:33
3
This post received
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

70% (02:20) correct 30% (02:21) wrong based on 65 sessions

### HideShow timer Statistics

A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

The official answer is:

[Reveal] Spoiler:
51/100

The official solution is the following:

[Reveal] Spoiler:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Positive outcomes = A + B - 2*C

Probability = 51/100

Can someone please explain why they subtract C twice?

Thank you.

Last edited by Bunuel on 07 Jul 2013, 05:43, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 38861
Followers: 7728

Kudos [?]: 106079 [8] , given: 11607

Re: Box with 100 tickets marked 1...100 [#permalink]

### Show Tags

07 Feb 2011, 03:24
8
This post received
KUDOS
Expert's post
13
This post was
BOOKMARKED
nonameee wrote:
The question from a textbook on probability:

A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

The official answer is:

[Reveal] Spoiler:
51/100

The official solution is the following:

[Reveal] Spoiler:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Positive outcomes = A + B - 2*C

Probability = 51/100

Can someone please explain why they subtract C twice?

Thank you.

A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

Total = 100 numbers.

Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Now, both A (numbers divisible by 2) and B (numbers divisible by 3) contain C (numbers divisible by 6) so to get the numbers which are divisible by either 2 or 3, but not 6 we should subtract C twice, once for A and once for B: A + B - 2*C= 51

Probability = 51/100

If it were: "What is the probability that the number on the ticket will be divisible by either 2 or 3 (or both)?" Then we would subtract C only once to get rid of double counting.
_________________
Director
Joined: 23 Apr 2010
Posts: 581
Followers: 2

Kudos [?]: 86 [0], given: 7

Re: Box with 100 tickets marked 1...100 [#permalink]

### Show Tags

07 Feb 2011, 04:12
Bunuel, thanks. I got it.
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2013
Followers: 163

Kudos [?]: 1822 [0], given: 376

Re: Box with 100 tickets marked 1...100 [#permalink]

### Show Tags

07 Feb 2011, 06:09
1
This post was
BOOKMARKED
I thought it to be 67/100 and realized my mistake.

Count of numbers divisible by 2 = 50. In these 50 numbers there are 16 numbers that are divisible by 6 and we must take them out. 50-16=34

Count of numbers divisible by 3 = 33. In these 33 numbers there are 16 numbers that are divisible by 6 and we must take them out. 33-16=17

Total numbers divisible by 2 or 3 but not 6 = 34+17=51

Probability = 51/100.

Good one.
_________________
Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 633
Followers: 44

Kudos [?]: 971 [0], given: 39

Re: Box with 100 tickets marked 1...100 [#permalink]

### Show Tags

08 Feb 2011, 12:11
Current Student
Joined: 06 Sep 2013
Posts: 2005
Concentration: Finance
Followers: 68

Kudos [?]: 643 [0], given: 355

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

25 Nov 2013, 08:26
nonameee wrote:
A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

The official answer is:

[Reveal] Spoiler:
51/100

The official solution is the following:

[Reveal] Spoiler:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Positive outcomes = A + B - 2*C

Probability = 51/100

Can someone please explain why they subtract C twice?

Thank you.

Every 6 numbers there are 3 divisible by 2 or 3 and not divisible by 6
Then in 100 numbers there are 16 sets of 6 and 4 numbers left as remainder.
So, 16(3) = 48. Now of the numbers remaining (92,98,99 and 100) there 3 of them are multiples of 2 or 3 but not 6.
So total is 51 numbers
Hence prob is 51/100

Hope it helps
Cheers
J
Intern
Joined: 13 Dec 2013
Posts: 40
Schools: Fuqua (I), AGSM '16
GMAT 1: 620 Q42 V33
Followers: 2

Kudos [?]: 18 [0], given: 10

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

01 Jun 2014, 08:59
Is it the same as saying: divisible by either 2 or 3 but not by both?
Math Expert
Joined: 02 Sep 2009
Posts: 38861
Followers: 7728

Kudos [?]: 106079 [0], given: 11607

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

01 Jun 2014, 09:49
Enael wrote:
Is it the same as saying: divisible by either 2 or 3 but not by both?

_______
Yes, it is the same.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15434
Followers: 649

Kudos [?]: 208 [0], given: 0

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

14 Jul 2015, 06:55
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.
_________________
Intern
Status: One more shot
Joined: 01 Feb 2015
Posts: 38
Location: India
Concentration: General Management, Finance
WE: Corporate Finance (Commercial Banking)
Followers: 0

Kudos [?]: 19 [0], given: 136

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

22 May 2016, 03:32
nonameee wrote:
A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

The official answer is:

[Reveal] Spoiler:
51/100

The official solution is the following:

[Reveal] Spoiler:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Positive outcomes = A + B - 2*C

Probability = 51/100

Can someone please explain why they subtract C twice?

Thank you.

51/100 wud be the soln
_________________

Believe you can and you are halfway there-Theodore Roosevelt

Director
Joined: 18 Oct 2014
Posts: 908
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
Followers: 98

Kudos [?]: 300 [1] , given: 69

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

22 May 2016, 10:59
1
This post received
KUDOS
nonameee wrote:
A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

The official answer is:

[Reveal] Spoiler:
51/100

The official solution is the following:

[Reveal] Spoiler:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Positive outcomes = A + B - 2*C

Probability = 51/100

Can someone please explain why they subtract C twice?

Thank you.

Total numbers= 100
numbers divisible by 2= 50
numbers divisible by three= 33
numbers divisible by 6= 16

numbers divisible by 2 and 3 but not 6= 50-16+33-16= 51

probability= 51/100
_________________

I welcome critical analysis of my post!! That will help me reach 700+

Intern
Joined: 09 Dec 2015
Posts: 27
Location: India
Concentration: General Management, Operations
GPA: 2.8
WE: Engineering (Manufacturing)
Followers: 0

Kudos [?]: 6 [0], given: 35

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

27 Aug 2016, 09:38
Please update the options for this question.
Correct Ans should be 51/100.
No. of multiples of 2 = 50
No. of multiples of 3 = 33
No. of multiples of 6 = 16
Since both multiples of 2 & 3 will also contain multiples of 6 so we need to subtract it twice.
Total no. of multiples of 2 & 3 but not 6 = 51.
Probability = 51/100
Manager
Joined: 20 Jan 2017
Posts: 65
Location: United States (NY)
GMAT 1: 750 Q48 V44
GMAT 2: 610 Q34 V41
Followers: 2

Kudos [?]: 5 [0], given: 15

Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]

### Show Tags

24 Jan 2017, 20:23
1) 100/2=50
2) 100/3=33
3) 100/3*2=16
4) 50-16+33-16=51
Re: A box contains 100 tickets marked 1 to 100. One ticket is   [#permalink] 24 Jan 2017, 20:23
Similar topics Replies Last post
Similar
Topics:
1 Tickets for all but 100 seats in a 10,000-seat stadium were sold. Of t 2 10 Jun 2016, 08:57
22 Jamboree and GMAT Club Contest: A box contains 100 balls, numbered 29 06 Feb 2017, 00:07
A bag contains 30 tickets, numbered from ‘1’ to ‘30’. Five tickets are 5 20 Dec 2011, 11:05
69 A box contains 100 balls, numbered from 1 to 100. If three b 28 20 Jun 2016, 03:42
A box contains 100 balls, numbered from 1 to 100. If three balls are 5 09 Aug 2010, 03:06
Display posts from previous: Sort by

# A box contains 100 tickets marked 1 to 100. One ticket is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.