A book contains 732 pages numbered from 1 to 732. If a student randoml

Question

A book contains 732 pages numbered from 1 to 732. If a student randomly opens the book, what is the probability that the page number contains the digit 1?

A. 112/732

B. 184/732

C. 227/732

D. 233/732

E. 254/732

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chetan2u · Answer

The probability depends on numbers of integers up to 732 containing digit 1.
Let us check how many digits in each set of 100 contain 1.
Units digit as 1.....
We can have the tens digit as any of the 10 digits 0-9 => 10*1=10
Tens digit as 1......
The ones digit can be any of the digits 0-9, so => 1*10=10
Subtract 1 as the common between above two...Both include the integer 11.
Total =10+10-1=19.
a) So 0-99, 200-299, 300-399, 400-499, 500-599 and 600-699 will contain 19 integers containing 1=> 19*6=114
b) Now 100-199 will have 1 in each integer => 100
c) 700-732....701,721,731 and 710-719 => 13

Total = 114+100+13=227

Probability of any of these 227 pages to be opened = 227/732

C

Kunni · Answer

Could you please tell me how did you calculate unite digit as 1 and tens digit as 1.

chetan2u · Answer

I have added further details in original post. Please have a look and ask if any further query.

ShankSouljaBoi · Answer

Hi  chetan2u  , 
could you please point out the flaw in my approach

Total outcomes = 732
Favourable outcomes = 
repition of 1 in any three place----- no of ways = 1
+
repition of 1 only in one place----- no of ways = 3(1*9*9) = 243
+
repition of 1 in two places----- no of ways = 3(1*1*9) = 27

= 271

Final answer = 271/732 ... i am not able to figure out what cases am i actually repeating

Best

chetan2u · Answer

You have found all the cases till 1000. So subtract the 1s from 732 to 1000.
19 each from 800-900 and 900-1000 and 6 from 732-800.
So 271-2*19-6=227.

carouselambra · Answer

Hi  chetansharma

Could you please help me here?
I was able to understand and work as per the the Odd/even strategy that you pointed out.
However, within abcd (Even), we can have multiple cases as shown.

Can you please point the flaw?
Thanks!

chetan2u · Answer

Firstly in all the individual cases the denominator will be 5^5 or 3125 and not 125.
Secondly when you calculate probability of 1odd and 3even, it becomes 4C1*(2/5)^3*3/5*3/5=4*72/3125
Similarly for others

somyamehta777 · Answer

Did this with counting technique -

1's in the hundreth digit = 1 * 10 * 10

1's in the tenth digit = 6*1*10 + 1*1*10
Here divided it into 2 parts - 1's in the tenth digit from 0-699 + 1's in the tenth digit from 700-732

1's in the unit digit = 6*9*1 + 3
Here divided it into 2 parts - 1's in units digit from 0-699 + 1's in the unit digit from 700-732(701, 711(but this is already calculated in the tenth digit), 721, 731 -> therefore, 3)

Total = 100 + 70 + 57 = 227

Probability = 227/732

akadiyan · Answer

A book contains 732 pages numbered from 1 to 732. If a student randomly opens the book, what is the probability that the page number contains the digit 1?

We can segregate this into 3 different sections

Case 1: 100 - 199 - 100 pages will have 1 in atleast one of the pages =  100

Case 2: 1 - 99 pages:  19  
Page 1-  1  page /  10  pages from 10 -19 /  8  pages in unit digits in 21,31....91

The above pattern will be the same for 200s to 600s , so we have 6*19 =  114  pages

Case 3: 700 - 732
Same calculation as case 2 , but the unit digits only till 732, so we have 1+10+2 =  13  pages

Adding all the cases together we get - 100+114+13 = 227

Total outcomes: 732
Favorable outcomes: 227

Probability: \frac{ 227}{732}

Ans:  C

HarshavardhanR · Answer

(1) Between 1 to 9 -> 1 page contains a 1 (page 1).
(2) Between 10 to 19 -> all pages contain a 1 i.e., 19-10+1 = 10 pages.
(3) Between 20 to 99 -> 21,31...91 i.e., (91-21)/10 +1 = 8 pages.
(4) Between 100 to 199 -> all pages contain a 1 i.e., 199-100+1 = 100 pages.

So, 
a) From page 1 to page 99 -> 1 + 10 + 8 = 19 pages. 
b) From page 100 to 199 -> 100 pages.
c) from page 200 to 299 -> (a) repeats. The only 1's are in the ten's digit and/or one's digit i.e., 201, then 210 to 219, then 221, 231...291. Thus, there are again 19 pages.

d) Overall, from page 200 to page 699 -> 5 sets of 100 pages, each with 19 pages having 1 -> 19 x 5 = 95.

e) from 699 to 732 -> 701 (1 page), 710 to 719 (10 pages), 721 and 731 (2 pages) -> 1+10+2 = 13 pages.

Total number of pages containing 1 = (a)+(b)+(d)+(e) = 19+100+95+13 = 227 pages.

Required Probability = 227/732. Choice C.
___
Harsha
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Kinshook · Answer

Given: A book contains 732 pages numbered from 1 to 732.

Asked: If a student randomly opens the book, what is the probability that the page number contains the digit 1?

Page numbers containing digit 1 = {1, 10,11,12,13,....19,21,31,41,51,61,71,81,91,100,101,110, 111,112,113,...119,120,121,..191..199,201,210,211,213,213,.219,221,231,.... 291,301,310,311, 312, ...319, 321,331,....391,401,410,411,412,413,..419,421,....491,501,510,511,512,513,...519,521,....591,601,610,611,612,613,..619,621,....691,701,710,711,712,713,..719,721,731}

Number of page numbers containing digit 1 = 2 + 9 + 8 + 100 + 2 + 9 + 8 + 2 + 9 + 8 + 2 + 9 + 8 + 2 + 9 + 8 + 2 + 9 + 8 + 2 + 9 + 2 = 19 + 100 + 19 + 19 + 19 + 19 + 19 + 13 = 227

The probability that the page number contains the digit 1 = 227/732

IMO C

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