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Probability Question- Expert needed

by Mo2men » Thu Feb 18, 2016 12:53 pm
A six-page spreadsheet is constructed so that each row displays one number, and all numbers are consecutive beginning at 1 (so line 1 contains the number 1; line 2 contains the number 2; etc.). If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?

(1) Each page has 72 rows of numbers.

(2) The number will be chosen from page 2.

Fact 2) No info about number of rows. so it is Insufficient.


However I'm confused in fact 1).

I used "Or" rule to determine probability.

Total prob= Prob of picking number from page 1 + prob to pick number from page 2.......+ Probability to pick from page 6

so Probability for one page (counting numbers divisible by 3)/72 = 24/72= 1/3

Probability for 6 pages= 1/3 * 6=2. I have not seen probability called 200%

where is my mistake? is it wrong to use 'or' rule?

Thanks
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by MartyMurray » Thu Feb 18, 2016 1:08 pm
Mo2men wrote:I'm confused in fact 1).

I used "Or" rule to determine probability.

Total prob= Prob of picking number from page 1 + prob to pick number from page 2.......+ Probability to pick from page 6

so Probability for one page (counting numbers divisible by 3)/72 = 24/72= 1/3

Probability for 6 pages= 1/3 * 6=2. I have not seen probability called 200%

where is my mistake? is it wrong to use 'or' rule?
Think about what you did. If you are going to take the number of pages into account, is the probability of picking a number that is

divisible by 3, and

on Page 1

really 1/3?

The are other ways to get to the answer, but as long as we are talking about your way, let's make your way work.
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by GMATGuruNY » Thu Feb 18, 2016 2:07 pm
Mo2men wrote:A six-page spreadsheet is constructed so that each row displays one number, and all numbers are consecutive beginning at 1 (so line 1 contains the number 1; line 2 contains the number 2; etc.). If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?

(1) Each page has 72 rows of numbers.

(2) The number will be chosen from page 2.
Statement 1:
Of every 3 rows, exactly 1 will contain an integer divisible by 3, as illustrated by the values in red:
1, 2, 3, 4, 5, 6...67, 68, 69, 70, 71, 72.
This reasoning will hold true for all 6 pages.
Thus, P(integer divisible by 3) = 1/3.
SUFFICIENT.

Statement 2:
If page 2 has 3 rows, then exactly 1 row will contain an integer divisible by 3, with the result that P(integer divisible by 3) = 1/3.
If page 2 has only 1 row -- and the integer in this row is NOT a multiple of 3 -- then P(integer divisible by 3) = 0.
Since the probability can be different values, INSUFFICIENT.

The correct answer is A.
Last edited by GMATGuruNY on Thu Feb 18, 2016 3:30 pm, edited 1 time in total.
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by Mo2men » Thu Feb 18, 2016 2:44 pm
GMATGuruNY wrote:
Mo2men wrote:A six-page spreadsheet is constructed so that each row displays one number, and all numbers are consecutive beginning at 1 (so line 1 contains the number 1; line 2 contains the number 2; etc.). If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?

(1) Each page has 72 rows of numbers.

(2) The number will be chosen from page 2.
Statement 1:
Of every 3 rows, exactly 1 will contain an integer divisible by 3, as illustrated by the values in red:
1, 2, 3, 4, 5, 6...67, 68, 69, 70, 71, 72.
This reasoning will hold true for all 6 pages.
Thus, P(integer divisible by 3) = 1/3.
SUFFICIENT.

Statement 2:
If page 2 has 3 rows, then exactly 1 row will contain an integer divisible 3, with the result that P(integer divisible by 3) = 1/3.
If page 2 has only 1 row -- and the integer in this row is NOT a multiple of 3 -- then P(integer divisible 3) = 0.
Since the probability can be different values, INSUFFICIENT.

The correct answer is A.
Thanks Mitch.

the original question asks for 'If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?'


what if the question is altered to 'If each page has the same number of rows, and a number is chosen at random from any of SIX pages of the document, what is the probability that the number will be divisible by 3?'

Does it make any difference? I think it should be different because the effect of different pages should be considered.
I'm confused to choose from one page and six pages.
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by MartyMurray » Thu Feb 18, 2016 3:14 pm
Mo2men wrote:the original question asks for 'If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?'

what if the question is altered to 'If each page has the same number of rows, and a number is chosen at random from any of SIX pages of the document, what is the probability that the number will be divisible by 3?'

Does it make any difference? I think it should be different because the effect of different pages should be considered.

I'm confused to choose from one page and six pages.
The change that you are talking about makes no difference.

If on each page there are 72 rows each containing 1 number, the probability of choosing a multiple of 3 is 1/3 whether you know which page you are choosing from or you are choosing from any random page.

Though the wording of the question conveys an ambiguous meaning, I think that all the wording of the question means is that you are choosing one number from one page.

All the same, let's say you were to choose a page and then you were to want to know the probability of choosing a multiple of 3 from the numbers on that page. The answer would be 1/3.

Meanwhile, since all of the pages contain the same numbers, if you were to instead choose a number randomly from all of the numbers on all of the pages, the probability of choosing a multiple of 3 would remain the same.

Page 1: contains numbers 1 - 72 The probability of choosing a multiple of 3 is 1/3.

Pages 1, 2, 3, 4, 5, and 6 contain six 1's, six 2's, six 3's and six of every number up to 72. Among all those numbers 1/3 are multiples of 3. So the probability of choosing a multiple of 3 from all the numbers on all the pages is still 1/3.

You could have 1000 pages, and the probability would still be the same.

Don't get stuck via thinking about formulas. Formulas can be useful, but before using them, consider what is actually going on in the scenario being described.
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