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

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nonameee
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A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   Updated on: 02 Jul 2023, 01:59
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Question Stats:

38% (01:57) correct 62% (02:01) wrong based on 241 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?

A. 83/100
B. 67/100
C. 51/100
D. 49/100
E. 33/100

The official answer is:

Show Spoiler
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.
ShowHide Answer
Official Answer
C

Originally posted by nonameee on 07 Feb 2011, 02:33.
Last edited by Bunuel on 02 Jul 2023, 01:59, edited 2 times in total.
Renamed the topic and edited the question.
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Bunuel
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   07 Feb 2011, 03:24
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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:

Show Spoiler
51/100


The official solution is the following:

Show 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.
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   07 Feb 2011, 06:09
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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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nonameee
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   07 Feb 2011, 04:12
Bunuel, thanks. I got it.
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   08 Feb 2011, 12:11
thanks to bunuel and fluke
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   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:

Show Spoiler
51/100


The official solution is the following:

Show 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 :)
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   01 Jun 2014, 08:59
Is it the same as saying: divisible by either 2 or 3 but not by both?
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   01 Jun 2014, 09:49
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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.
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   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:

Show Spoiler
51/100


The official solution is the following:

Show 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
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   22 May 2016, 10:59
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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:

Show Spoiler
51/100


The official solution is the following:

Show 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
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   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
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   24 Jan 2017, 20:23
1) 100/2=50
2) 100/3=33
3) 100/3*2=16
4) 50-16+33-16=51
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   08 Aug 2019, 15:34
We can use a venn diagram for this

Divisible by both 2 and 3 means

Divisible by 6
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   28 Dec 2021, 11:26
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:

Show Spoiler
51/100


The official solution is the following:

Show 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.


Numbers divisible by 2
=100/2= 50
Numbers divisible by 3
=100/3= 33
Numbers divisible by 6
= 100/6=16
Now,Numbers divisible by 2 but not by 6= 50-16=34

And,Numbers divisible by 3 but not by 6= 33-16=17

Therefore,Numbers divisible by 2 and 3 but not by 6= 34+17=51

Probability = 51/100=0.51

Posted from my mobile device
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   17 Jan 2022, 08:04
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Total numbers= 100
Numbers divisible by 2 = 50
Numbers divisible by 3 = 33
Numbers divisible by 6 = 16

(Refer to the attached screenshot for the venn diagram)

Numbers divisible by either 2 or 3 but not 6 = 34+17 = 51

Probability = Favorable options/Total options

Favorable options = 51C1 = 51
Total options = 100C1 = 100

P(Number divisible by either 2 or 3 but not 6) = 51/100
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink] New post   17 Sep 2023, 07:44
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Re: A box contains 100 tickets marked 1 to 100. One ticket is [#permalink]
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