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

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07 Feb 2011, 02:33

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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 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?

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.
_________________

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

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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?

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

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

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14 Jul 2015, 06:55

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

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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?

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

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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?

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

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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

gmatclubot

Re: A box contains 100 tickets marked 1 to 100. One ticket is
[#permalink]
27 Aug 2016, 09:38

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