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Director  Joined: 23 Apr 2010
Posts: 527
A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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4
43 00:00

Difficulty:

(N/A)

Question Stats: 55% (01:15) correct 45% (02:18) wrong based on 126 sessions

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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:

51/100

The official solution is the following:

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.

Originally posted by nonameee on 07 Feb 2011, 02:33.
Last edited by Bunuel on 07 Jul 2013, 05:43, edited 1 time in total.
Renamed the topic and edited the question.
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 57298
Re: Box with 100 tickets marked 1...100  [#permalink]

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14
16
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:

51/100

The official solution is the following:

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.
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Retired Moderator Joined: 20 Dec 2010
Posts: 1707
Re: Box with 100 tickets marked 1...100  [#permalink]

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4
2
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.
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##### General Discussion
Director  Joined: 23 Apr 2010
Posts: 527
Re: Box with 100 tickets marked 1...100  [#permalink]

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Bunuel, thanks. I got it.
Senior Manager  S
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 472
Re: Box with 100 tickets marked 1...100  [#permalink]

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SVP  Joined: 06 Sep 2013
Posts: 1625
Concentration: Finance
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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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:

51/100

The official solution is the following:

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: 36
GMAT 1: 620 Q42 V33 Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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Is it the same as saying: divisible by either 2 or 3 but not by both?
Math Expert V
Joined: 02 Sep 2009
Posts: 57298
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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Enael wrote:
Is it the same as saying: divisible by either 2 or 3 but not by both?

_______
Yes, it is the same.
_________________
Intern  S
Status: One more try
Joined: 01 Feb 2015
Posts: 42
Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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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:

51/100

The official solution is the following:

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
_________________
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Current Student Joined: 18 Oct 2014
Posts: 826
Location: United States
GMAT 1: 660 Q49 V31 GPA: 3.98
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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1
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:

51/100

The official solution is the following:

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+
Current Student S
Joined: 09 Dec 2015
Posts: 113
Location: India
Concentration: General Management, Operations
Schools: IIMC (A)
GMAT 1: 700 Q49 V36 GPA: 3.5
WE: Engineering (Consumer Products)
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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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
Current Student B
Joined: 20 Jan 2017
Posts: 56
Location: United States (NY)
Schools: CBS '20 (A)
GMAT 1: 750 Q48 V44 GMAT 2: 610 Q34 V41 GPA: 3.92
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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1) 100/2=50
2) 100/3=33
3) 100/3*2=16
4) 50-16+33-16=51
Intern  B
Joined: 06 May 2019
Posts: 27
Location: India
Concentration: General Management, Marketing
Re: A box contains 100 tickets marked 1 to 100. One ticket is  [#permalink]

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We can use a venn diagram for this

Divisible by both 2 and 3 means

Divisible by 6 Re: A box contains 100 tickets marked 1 to 100. One ticket is   [#permalink] 08 Aug 2019, 15:34
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