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Probability Problem

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by Sak32 » Wed Nov 06, 2013 2:30 am
A book contains 732 pages numbered 1,2....732. If a student randomly opens the book, whats is the probability that the page number contains digit 1?

I had 246/732 for an answer and wanted to make sure its the right answer.
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by ganeshrkamath » Wed Nov 06, 2013 2:54 am
Sak32 wrote:A book contains 732 pages numbered 1,2....732. If a student randomly opens the book, whats is the probability that the page number contains digit 1?

I had 246/732 for an answer and wanted to make sure its the right answer.
0xy : 0x1, 01x => 10 + 10 - 1 = 19
1xy : 100
2xy : 19
3xy : 19
4xy : 19
5xy : 19
6xy : 19
700 to 732: 1 + 10 + 2 = 13

Total = 19*7 + 100 + 13 = 133 + 100 + 13
= 246

EDIT:
Total = 19*6 + 100 + 13 = 114 + 100 + 13
= 227

Probability = 227/732

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Last edited by ganeshrkamath on Wed Nov 06, 2013 11:12 am, edited 1 time in total.
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by theCodeToGMAT » Wed Nov 06, 2013 2:58 am
Pages 732

_ _ _

1 in Unit digit = 730/10 = 73 + 1 = 74

1 in Ten's Digit = 700/10 = 70 + 10 = 80

1 in Hundred's Digit = 100

So total = 100 + 74 + 80 = 254

So, 254/732
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by ganeshrkamath » Wed Nov 06, 2013 3:03 am
theCodeToGMAT wrote:Pages 732

_ _ _

1 in Unit digit = 730/10 = 73 + 1 = 74

1 in Ten's Digit = 700/10 = 70 + 10 = 80

1 in Hundred's Digit = 100

So total = 100 + 74 + 80 = 254

So, 254/732
Remove x11 from the total
254 - 8 = 246

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by theCodeToGMAT » Wed Nov 06, 2013 3:41 am
I believe the answer must be 227/732 because we need probability of "1" in selected number.. so, we must consider total number of numbers which contain "1"..

So,

0 - 99 ==> _ _ ==> 10 + 10 = 20
100 - 200 ==> _ _ _ ==> 100 + 10 + 10 = 120
201 - 300 ==> _ _ _ ==> 0 + 10 + 10
301 - 400 ==> _ _ _ ==> 0 + 10 + 10
401 - 500 ==> _ _ _ ==> 0 + 10 + 10
501 - 600 ==> _ _ _ ==> 0 + 10 + 10
601 - 700 ==> _ _ _ ==> 0 + 10 + 10
701 - 732 ==> _ _ _ ==> _ + 10 + 4

Occurrence of _11 = 7 (11,211,311,411,511,611,711)
Occurrence of 1_1 = 9 (101,121,131,141,151,161,171,181,191)
Occurrence of 11_ = 9 (110,112,113......119)
occurrence of 111 = 1 (we need to do -2)

So, 254 - 7 - 9 - 9 -2 = 227

227/732
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by GMATGuruNY » Wed Nov 06, 2013 4:01 am
Sak32 wrote:A book contains 732 pages numbered 1,2....732. If a student randomly opens the book, whats is the probability that the page number contains digit 1?

I had 246/732 for an answer and wanted to make sure its the right answer.
Pages between 000 and 699, inclusive, that do NOT include 1:
Number of options for the hundreds place = 6. (Any digit 0-6 but 1.)
Number of options for the tens place = 9. (Any digit 0-9 but 1.)
Number of options for the units place = 9. (Any digit 0-9 but 1.)
To combine these options, we multiply:
6*9*9 = 486.

Pages between 000 and 699 that DO include 1:
Since 486 of the 700 pages do NOT include 1, the number of pages that DO include 1 = 700-486 = 214.

Pages between 700 and 732, inclusive, that include 1:
701
710 - 719
721
731
Total = 13.

Thus:
Total number of pages that include 1 = 214+13 = 227.
P(page includes 1) = 227/732.
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by ganeshrkamath » Wed Nov 06, 2013 11:13 am
ganeshrkamath wrote:Probability = 227/732
Edited above.

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